Mechanics Engineering - Statics
Chapter 8
NB T = Find ( N , T , m , x) B D mD x
Since x = 51.6 mm < b 2
NB 413.05 = N T 129.08
mD = 25.6 kg x = 0.052 m
= 125 mm our assumption is correct
mD = 25.6 kg
Problem 8-106 Block A rests on the surface for which the coefficient of friction is sAB. If the mass of the suspended cylinder is mD, determine the smallest mass mA of block A so that it does not slip or tip. The coefficient of static friction between the cord and the fixed peg at C is sC. Units Used: g = 9.81 m s Given:
2
sAB = 0.25
mD = 4 kg
sC = 0.3
a = 0.3 m b = 0.25 m c = 0.4 m d = 3 f = 4 Solution: Assume that slipping is the crtitical motion NB = 100 N T = 50 N f = - atan
d
The initial guesses:
mA = 1 kg
x = 10 mm
Given
sC mD g = T e
d 2 2 T - mA g + NB = 0 f +d
865
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
- f T + N = 0 sAB B 2 2 f +d NB T = Find ( N , T , m , x) B A mA x
Since x = 51.6 mm < b 2
f d b 2 2 T a - 2 2 T 2 - NB x = 0 f +d f +d
NB 64.63 = N T 20.20
mA = 7.82 kg x = 0.052 m mA = 7.82 kg
= 125 mm our assumption is correct
Problem 8-107 The collar bearing uniformly supports an axial force P. If the coefficient of static friction is s, determine the torque M required to overcome friction. Given: a = 2 in b = 3 in P = 800 lb
s = 0.3
Solution:
b 3 a 3 - a 2 2 M = s P 2 b b 2 - a 2 2
M =
1 a s P b 2
a2 + a b + b2 a + b
M = 304.00 lb in
Problem 8-108 The collar bearing uniformly supports an axial force P. If a torque M is applied to the shaft and causes it to rotate at constant velocity, determine the coefficient of kinetic friction at the surface of contact.
866
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given: a = 2 in b = 3 in P = 500 lb M = 3 lb ft
Solution:
b 3 a 3 - 3 3 a 2 2 = a k P b - a M = k P 2 2b 2 2 b b 2 a b - a - 2 2
k =
2M b b - a aP b - a
(3
(2
2
3
)
)
k = 0.0568
Problem 8-109 The double-collar bearing is subjected to an axial force P . Assuming that collar A supports kP and collar B supports (1 - k)P, both with a uniform distribution of pressure, determine the maximum frictional moment M that may be resisted by the bearing. Units Used: kN = 10 N Given: P = 4 kN a = 20 mm b = 10 mm c = 30 mm
3
s = 0.2
867
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
k = 0.75 Solution: M =
2 c - b a -b s kP + ( 1 - k)P 2 2 2 3 2 a -b c - b
3 3 3 3
M = 16.1 N m
Problem 8-110 The annular ring bearing is subjected to a thrust P. If the coefficient of static friction is s, determine the torque M that must be applied to overcome friction. Given: P = 800 lb
s = 0.35
a = 0.75 in b = 1 in c = 2 in Solution:
b - c 2 M = s P b2 - c2 3
3 3
M = 36.3 lb ft
Problem 8-111 The floor-polishing machine rotates at a constant angular velocity. If it has weight W, determine the couple forces F the operator must apply to the handles to hold the machine stationary. The coefficient of kinetic friction between the floor and brush is k. Assume the brush exerts a uniform pressure on the floor. Given: W = 80 lb
k = 0.3
868
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
a = 1.5 ft b = 2 ft Solution: Fa = 2 b k W 3 2 1 k W 3
F =
b a
F = 10.7 lb
Problem 8-112 The plate clutch consists of a flat plate A that slides over the rotating shaft S. The shaft is fixed to the driving plate gear B. If the gear C, which is in mesh with B, is subjected to a torque M, determine the smallest force P, that must be applied via the control arm, to stop the rotation. The coefficient of static friction between the plates A and D is s. Assume the bearing pressure between A and D to be uniform. Given: M = 0.8 N m
s = 0.4
a = 150 mm b = 200 mm c = 100 mm d = 125 mm e = 150 mm f = 30 mm Solution: F = M f F = 26.667 N M2 = 4.00 N m
M2 = F e
869
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
d - c 2 M2 = s P' 3 d2 - c2
3 3
P' =
3M2 d2 - c2 2 s d3 - c3
P' = 88.5 N
P' b - P a = 0
P = P'
b a
P = 118 N
Problem 8-113 The shaft of diameter b is held in the hole such that the normal pressure acting around the shaft varies linearly with its depth as shown. Determine the frictional torque that must be overcome to rotate the shaft. Given: a = 6 in p0 = 60 b = 4 in lb in
2
s = 0.2
Solution: N =
a
x p 2 b dx 0 a 2
T = s N
b 2
T = 905 lb in
0
Problem 8-114 Because of wearing at the edges, the pivot bearing is subjected to a conical pressure distribution at its surface of contact. Determine the torque M required to overcome friction and turn the shaft, which supports an axial force P. The coefficient of static friction is s. For the solution, it is necessary to determine the peak pressure p0 in terms of P and the bearing radius R. Solution: P=
2
0
R
2 p - p r r dr d = p0 R 0 0 R 3
p0 =
3P
0
R
2
dM = rdF = r dN = r pdA = r p rd dr
870
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
p0 2 2 3P 3 P R 3 M= p0 - r r dr 1 d = p0 R = R = 2 2 R 6 6 0 R
0
R
Thus,
M=
P R
2
Problem 8-115 The conical bearing is subjected to a constant pressure distribution at its surface of contact. If the coefficient of static friction is s, determine the torque M required to overcome friction if the shaft supports an axial force P. Solution: The differential Area (shaded)
dA = 2r
dr = 2 rdr cos ( ) cos ( )
p cos ( )
P=
p cos ( ) d A =
2 r dr = 2 p r dr = p R2 0 cos ( )
R
p=
P
R
2
871
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
dN = pdA =
2P 2 rdr = rdr 2 cos ( ) 2 ( ) R R cos P
M = r dF = r s dN =
2 s P 2 2 r dr R cos ( ) 0
R
M=
2 s P R3 2 s P R = 2 ( ) 3 3 cos ( ) R cos
Problem 8-116 The tractor is used to push the pipe of weight W. To do this it must overcome the frictional forces at the ground, caused by sand. Assuming that the sand exerts a pressure on the bottom of the pipe as shown, and the coefficient of static friction between the pipe and the sand is s, determine the force required to push the pipe forward. Also, determine the peak pressure p0. Given: W = 1500 lb
s = 0.3
L = 12 ft r = 15 in
872
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
2 2L p0 cos ( ) r d cos ( ) - W = 0
0
+
Fy = 0;
2 2 W = 2 p0 L r cos d = p0 L r 0 2 p0 = 2 W rL p0 = 0.442 lb in
2
+ Fx = 0;
2 F = s p0 L r cos ( ) d -
2
F = 573 lb
Problem 8-117 Assuming that the variation of pressure at the bottom of the pivot bearing is defined as p = p0(R2/r), determine the torque M needed to overcome friction if the shaft is subjected to an axial force P. The coefficient of static friction is s. For the solution, it is necessary to determine p0 in terms of P and the bearing dimensions R 1 and R2. Solution: P= 0
2
R
R2
p r dr d
1
873
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
P=
2
0
R
R2
p0
1
R2 r dr d r
P = 2 p0 R2 ( R 2 - R 1 ) P
p0 =
2 R 2 ( R 2 - R 1 )
2
M = r dF = 0 A M=
2
R
R2
r s p0 r dr d
1
0
2
s p0
R2 2 2 2 r dr d = s p0 R 2 R 2 - R 1 r
(
)
2
0
P 2 2 M = s R2 R2 - R1 2 R2( R2 - R1 ) Problem 8-118
(
)
M=
s P(R2 + R1)
A disk having an outer diameter a fits loosely over a fixed shaft having a diameter b. If the coefficient of static friction between the disk and the shaft is s, determine the smallest vertical force P , acting on the rim, which must be applied to the disk to cause it to slip over the shaft. The disk weighs W. Given: a = 8 in b = 3 in
s = 0.15
W = 10 lb Solution:
s = atan ( s)
874
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
rf =
b sin ( ) s 2 a - r f 2
l =
M0 = 0; W rf - l P = 0 P = W
rf l
P = 0.59 lb
Problem 8-119 The pulley has a radius r and fits loosely on the shaft of diameter d. If the loadings acting on the belt cause the pulley to rotate with constant angular velocity, determine the frictional force between the shaft and the pulley and compute the coefficient of kinetic friction. The pulley has weight W. Given: r = 3 in d = 0.5 in W = 18 lb F 1 = 5 lb F 2 = 5.5 lb Solution:
+
Fy = 0;
MO = 0;
R - W - F1 - F2 = 0
R = W + F1 + F2 rf = r
R = 28.5 lb
-F 2 r + F1 r + R rf = 0 d sin ( k) 2
F2 - F1 R rf d
rf = 0.05263 in
rf =
k = asin 2 = tan ( k)
k = 12.15 deg = 0.215
Also, MO = 0; -F 2 r + F1 r + F
d = 0 2
F = 2 r
F2 - F1 d
F = 6 lb
875
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-120 The pulley has a radius r and fits loosely on the shaft of diameter d. If the loadings acting on the belt cause the pulley to rotate with constant angular velocity, determine the frictional force between the shaft and the pulley and compute the coefficient of kinetic friction. Neglect the weight of the pulley . Given: r = 3 in d = 0.5 in W = 0 lb F 1 = 5 lb F 2 = 5.5 lb Solution:
+
Fy = 0;
MO = 0;
R - W - F1 - F2 = 0 -F 2 r + F1 r + R rf = 0 d sin ( k) 2
R = W + F1 + F2 rf = r
R = 10.5 lb rf = 0.14286 in
F2 - F1 R rf d
rf =
k = asin 2 = tan ( k)
k = 34.85 deg = 0.696
Also, MO = 0; -F 2 r + F1 r + F d =0 2 F = 2 r
F2 - F1 d
F = 6 lb
Problem 8-121 A pulley of mass M has radius a and the axle has a diameter D. If the coefficient of kinetic friction between the axle and the pulley is k determine the vertical force P on the rope required to lift the block of mass MB at constant velocity. Given: a = 120 mm M = 5 kg D = 40 mm
876
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
k = 0.15
MB = 80 kg
Solution:
k = atan ( k)
rf =
D sin ( ) k 2
Mp = 0; MB g( a + rf) + M g rf - P( a - rf) = 0 MB g( a + rf) + M g rf a - rf
P =
P = 826 N
Problem 8-122 A pulley of mass M has radius a and the axle has a diameter D. If the coefficient of kinetic friction between the axle and the pulley is k determine the force P on the rope required to lift the block of mass MB at constant velocity. Apply the force P horizontally to the right (not as shown in the figure). Given: a = 120 mm M = 5 kg D = 40 mm
k = 0.15
MB = 80 kg Solution: g = 9.81 m s
2
k = atan ( k)
rf = D sin ( k) 2
877
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Guesses P = 1N Given R cos ( ) - MB g - M g = 0 P - R sin ( ) = 0 MB g a - P a + R rf = 0 R = 1N
= 1 deg
P R = Find ( P , R , )
P = 814 N
Problem 8-123 A wheel on a freight car carries a load W. If the axle of the car has a diameter D, determine the horizontal force P that must be applied to the axle to rotate the wheel. The coefficient of kinetic friction is k. Units Used: kip = 1000 lb Given: W = 20 kip D = 2 in
k = 0.05
r = 16 in Solution:
F x = 0; F y = 0;
P - R sin ( ) = 0 R cos ( ) - W = 0 P = W tan ( )
Thus,
k = atan ( k)
k = 2.86 deg
878
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
rf =
D sin ( ) k 2 rf r
rf = 0.04994 in
= asin
P = W tan ( ) P = 62.4 lb
= 0.1788 deg
Problem 8-124 The trailer has a total weight W and center of gravity at G which is directly over its axle. If the axle has a diameter D, the radius of the wheel is r, and the coefficient of kinetic friction at the bearing is k, determine the horizontal force P needed to pull the trailer. Given: W = 850 lb r = 1.5 ft
k = 0.08
D = 1 in Solution:
F x = 0; F y = 0;
P - R sin ( ) = 0 R cos ( ) - W = 0 P = W tan ( )
Thus,
k = atan ( k)
rf =
k = 4.57 deg
rf = 0.03987 in
D sin ( ) k 2 rf r
= asin
P = W tan ( )
= 0.1269 deg
P = 1.88 lb
879
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-125 The collar fits loosely around a fixed shaft that has radius r. If the coefficient of kinetic friction between the shaft and the collar is k, determine the force P on the horizontal segment of the belt so that the collar rotates counterclockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt is R.
Given: r = 2 in
k = 0.3
R = 2.25 in F = 20 lb
Solution:
k = atan ( k)
rf = r sin ( k)
k = 16.699 deg
rf = 0.5747 in
Equilibrium:
+
Fy = 0;
Ry - F = 0 P - Rx = 0 R=
2 2
Ry = F Rx = P P +F
2 2
R y = 20.00 lb
+ Fx = 0;
Rx + Ry =
Guess Given
P = 1 lb -
(
P +F
2
2
) rf + F R - P R = 0
P = Find ( P)
P = 13.79 lb
Problem 8-126 The collar fits loosely around a fixed shaft that has radius r. If the coefficient of kinetic friction between the shaft and the collar is k, determine the force P on the horizontal segment
880
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt is R.
Given: r = 2 in
k = 0.3
R = 2.25 in F = 20 lb
Solution:
k = atan ( k)
rf = r sin ( k)
k = 16.699 deg
rf = 0.5747 in
Equilibrium:
Fy = 0; + Fx = 0;
+
Ry - F = 0 P - Rx = 0 R=
2 2
Ry = F Rx = P P +F
2 2
R y = 20.00 lb
Rx + Ry =
Guess Given
P = 1 lb
(
P + F rf + F R - P R = 0
2
2
)
P = Find ( P)
P = 29.00 lb
Problem 8-127 The connecting rod is attached to the piston by a pin at B of diameter d1 and to the crank shaft by a bearing A of diameter d2. If the piston is moving downwards, and the coefficient of static friction at these points is s, determine the radius of the friction circle at each connection. Given: d1 = 0.75 in
881
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
d2 = 2 in
s = 0.2
Solution: rfA = 1 2 1 2 d2 s rfA = 0.2 in
rfB =
d1 s
rfB = 0.075 in
Problem 8-128 The connecting rod is attached to the piston by a pin at B of diameter d1 and to the crank shaft by a bearing A of diameter d2. If the piston is moving upwards, and the coefficient of static friction at these points is s, determine the radius of the friction circle at each connection. Given: d1 = 20 mm d2 = 50 mm
s = 0.3
Solution: rfA = rfB = 1 2 1 2 d2 s d1 s rfA = 7.50 mm rfB = 3 mm
882
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-129 The lawn roller has mass M. If the arm BA is held at angle from the horizontal and the coefficient of rolling resistance for the roller is r, determine the force P needed to push the roller at constant speed. Neglect friction developed at the axle, A, and assume that the resultant force P acting on the handle is applied along arm BA. Given: M = 80 kg
= 30 deg
a = 250 mm r = 25 mm Solution: r 1 = asin
a
1 = 5.74 deg
M0 = 0;
-r M g - P sin ( ) r + P cos ( ) a cos ( 1 ) = 0 rM g
P =
-sin ( ) r + cos ( ) a cos ( 1 )
P = 96.7 N
Problem 8-130 The handcart has wheels with a diameter D. If a crate having a weight W is placed on the cart, determine the force P that must be applied to the handle to overcome the rolling resistance. The coefficient of rolling resistance is . Neglect the weight of the cart.
883
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given: W = 1500 lb D = 6 in a = 0.04 in c = 3 b = 4
Solution: Guesses Given N - W - P c 2 2 = 0 c +b N = 1 lb P = 1 lb
b P = N 2a 2 2 D b +c N = Find ( N , P) P
N = 1515 lb P = 25.3 lb
Problem 8-131 The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for the cylinder's top and bottom surfaces are aA and aB respectively, show that a force having a magnitude of P = [W(aA + aB)]/2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder.
884
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
+ Fx = 0; +
R Ax - P = 0
R Ax = P
Fy = 0;
MB = 0;
R Ay - W = 0
R Ay = P
P ( r cos ( A) + r cos ( B) ) - W ( aA + aB) = 0
Since and B are very small, cos ( A) = cos ( B) = 1 Hence from Eq.(1) W ( aA + aB) 2r
P=
(QED)
Problem 8-132 A steel beam of mass M is moved over a level surface using a series of rollers of diameter D for which the coefficient of rolling resistance is ag at the ground and as at the bottom surface of the beam. Determine the horizontal force P needed to push the beam forward at a constant speed. Hint: Use the result of Prob. 8131. Units Used: Mg = 1000 kg
885
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given: M = 1.2 Mg D = 30 mm ag = 0.4 mm as = 0.2 mm Solution: P = M g( ag + as) D 2 2 P = 235 N
Problem 8-133 A machine of mass M is to be moved over a level surface using a series of rollers for which the coefficient of rolling resistance is ag at the ground and am at the bottom surface of the machine. Determine the appropriate diameter of the rollers so that the machine can be pushed forward with a horizontal force P. Hint: Use the result of Prob. 8-131. Units Used: Mg = 1000 kg Given: M = 1.4 Mg ag = 0.5 mm am = 0.2 mm P = 250 N Solution: P= M g( ag + am) 2r
r = M g
ag + am 2P
r = 19.2 mm
d = 2r
d = 38.5 mm
886
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-134 A single force P is applied to the handle of the drawer. If friction is neglected at the bottom and the coefficient of static friction along the sides is s determine the largest spacing s between the symmetrically placed handles so that the drawer does not bind at the corners A and B when the force P is applied to one of the handles. Given:
s = 0.4
a = 0.3 m b = 1.25 m Solution: Equation of Equilibrium and Friction : If the drawer does not bind at corners A and B, slipping would have to occur at points A and B. Hence, FA = NA and FB = NB
+ Fx = 0;
NB - NA = 0 NA = NB = N
+
Fy = 0;
MB = 0;
s NA + s NB - P = 0
N a + s N b - P
P = 2 s N
s + b = 0 2
a + b - 2 s + b N = 0 s s 2
a + s b - s ( s + b) = 0 s = a
s
s = 0.750 m
Problem 8-135 The truck has mass M and a center of mass at G. Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has
887
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
( ) four-wheel drive. The coefficient of static friction between the wheels and the ground is st and between the crate and the ground, it is sc. Units Used: kN = 10 N Mg = 1000 kg Given: M = 1.25 Mg
3
st = 0.5 sc = 0.4
g = 9.81 Solution: Guesses T = 1N m s
2
a = 600 mm b = 1.5 m c = 1m d = 800 mm
NA = 1 N NC = 1 N
NB = 1 N W = 1N
(a) Rear wheel drive Given -T + st NA = 0 NA + NB - M g = 0 -M g b + NB( b + c) + T a = 0 T - sc NC = 0 NC - W = 0
NA NB T = Find ( N , N , T , N , W) A B C NC W
(b) Four wheel drive Given -T + st NA + st NB = 0
W = 6.97 kN
888
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
NA + NB - M g = 0 -M g b + NB( b + c) + T a = 0 T - sc NC = 0 NC - W = 0
NA NB T = Find ( N , N , T , N , W) A B C NC W
W = 15.33 kN
Problem 8-136 The truck has M and a center of mass at G. The truck is traveling up an incline of angle . Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has four-wheel drive. The coefficient of static friction between the wheels and the ground is st and between the crate and the ground, it is sc.
Units Used: kN = 10 N Mg = 1000 kg Given:
3
= 10 deg
889
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
M = 1.25 Mg a = 600 mm
st = 0.5 sc = 0.4
g = 9.81 Solution: Guesses T = 1N m s
2
b = 1.5 m c = 1m d = 800 mm
NA = 1 N NC = 1 N
NB = 1 N W = 1N
(a) Rear wheel drive Given -T + st NA - M g sin ( ) = 0 NA + NB - M g cos ( ) = 0 -M g b cos ( ) + M g d sin ( ) + NB( b + c) + T a = 0
T - sc NC - W sin ( ) = 0 NC - W cos ( ) = 0
NA NB T = Find ( N , N , T , N , W) A B C NC W
(b) Four wheel drive Given
W = 1.25 kN
-T + st NA + st NB - M g sin ( ) = 0 NA + NB - M g cos ( ) = 0 -M g b cos ( ) + M g d sin ( ) + NB( b + c) + T a = 0 T - sc NC - W sin ( ) = 0 NC - W cos ( ) = 0
890
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
NA NB T = Find ( N , N , T , N , W) A B C NC W
Problem 8-137
W = 6.89 kN
A roofer, having a mass M, walks slowly in an upright position down along the surface of a dome that has a radius of curvature . If the coefficient of static friction between his shoes and the dome is s determine the angle at which he first begins to slip. Given: M = 70 kg
= 20 m s = 0.7
Solution:
F y' = 0; F x' = 0;
Nm - M g cos ( ) = 0 M g sin ( ) - s Nm = 0
s = tan ( ) = atan ( s) = 35.0 deg
Problem 8-138 A man attempts to lift the uniform ladder of weight W to an upright position by applying a force P perpendicular to the ladder at rung R. Determine the coefficient of static friction between the ladder and the ground at A if the ladder begins to slip on the ground when his
891
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
g hands reach height c. Given: a = 2 ft b = 14 ft c = 6 ft W = 40 lb Solution: c = asin
g
p
g
b
Initial guesses P = 10 lb Given F x = 0; F y = 0; MA = 0; NA = 100 lb
A = 100
A NA - P sin ( ) = 0
NA - W + P cos ( ) = 0 -W
cos ( ) + P b = 0 2
b + a
P NA = Find ( P , NA , A) A
P 20.7 = lb NA 21.3
A = 0.41
Problem 8-139 Column D is subjected to a vertical load W. It is supported on two identical wedges A and B for which the coefficient of static friction at the contacting surfaces between A and B and between B and C is s. Determine the force P needed to raise the column and the equilibrium force P' needed to hold wedge A stationary. The contacting surface between A and D is smooth.
892
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Units Used: kip = 10 lb Given: W = 8000 lb
3
= 10 deg s = 0.4
Solution: wedge A: F y = 0; N cos ( ) - s N sin ( ) - W = 0 N = cos ( ) - s sin ( ) W
N = 8739.8 lb F x = 0;
s N cos ( ) + N sin ( ) - P' = 0
P' = s N cos ( ) + N sin ( ) P' = 4.96 kip
Wedge B: F y = 0; NC + s N sin ( ) - N cos ( ) = 0 NC = - s N sin ( ) + N cos ( ) NC = 8000 lb F x = 0; P - s NC - N sin ( ) - s N cos ( ) = 0 P = s NC + N sin ( ) + s N cos ( ) P = 8.16 kip Problem 8-140 Column D is subjected to a vertical load W. It is supported on two identical wedges A and B for which the coefficient of static friction at the contacting surfaces between A and B and between B and C is s. If the forces P and P' are removed, are the wedges self-locking? The contacting surface between A and D is smooth.
893
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given: W = 8000 lb
= 10 deg s = 0.4
Solution: Wedge A: F y = 0; N - W cos ( ) = 0 N = W cos ( ) N = 7878.5 lb F x = 0; W sin ( ) - F = 0 F = W sin ( ) F = 1389.2 lb Friction F max = s N
Since F = 1389 lb < Fmax = 3151 lb then the wedges do not slip at the contact surface AB. Wedge B: F y = 0; NC - F sin ( ) - N cos ( ) = 0 NC = F sin ( ) + N cos ( ) NC = 8000 lb F x = 0; F C + F cos ( ) - N sin ( ) = 0 F C = -F cos ( ) + N sin ( ) F C = 0 lb Friction F Cmax = s NC
Since F C = 0 lb < FCmax = 3200 lb then the wedges do not slip at the contact surface BC. Therefore the wedges are self-locking.
894
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-1 Locate the center of mass of the homogeneous rod bent in the form of a parabola. Given: a = 1m b = 2m Solution:
x y = b a
2
d y 2b = x 2 dx a 2 2 x 1 + 2b x dx b 2 a a
0 a
yc =
0
a
yc = 0.912 m
xc = 0 m
2b 1 + x dx 2 a
2
Problem 9-2 Locate the center of gravity xc of the homogeneous rod. If the rod has a weight per unit length , determine the vertical reaction at A and the x and y components of reaction at the pin B. Given:
= 0.5
a = 1 ft b = 2 ft
lb ft
895
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: y = b
x a
a
2
d y 2b = x 2 dx a
L = 0
2 b 1+ 2 a
a
x dx
2
W = L
W = 1.162 lb
2 1 2 b x dx xc = x 1+ 2 L a 0 Guesses Given Ay = 1lb Bx = 0 B x = 1lb Ay + By - W = 0
xc = 0.620 ft
B y = 1lb - Ay a + W ( a - xc) = 0
Bx By = Find ( Bx , By , Ay) A y
Bx 0.000 By = 0.720 lb A 0.442 y
Problem 9-3 Locate the center of mass of the homogeneous rod bent into the shape of a circular arc. Given: r = 300 mm
= 30 deg
Solution: yc = 0 Symmetry
896
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
+ 2 r cos ( ) r d - -
xc =
2
+ 2 r d - - 2
xc = 124.049 mm
Problem 9-4 Locate the center of gravity xc of the homogeneous rod bent in the form of a semicircular arc. The rod has a weight per unit length . Also, determine the horizontal reaction at the smooth support B and the x and y components of reaction at the pin A. Given:
= 0.5
r = 2 ft
lb ft
Solution:
2 r cos ( ) r d - xc =
2
2 r d -
2
xc = 1.273 ft
MA = 0;
- r xc + B x ( 2 r) = 0
Bx =
r xc
2r
B x = 1 lb
897
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
+ Fx = 0; +
- Ax + Bx = 0 Ay - r = 0
Ax = Bx Ay = r
Ax = 1 lb Ay = 3.14 lb
Fy = 0;
Problem 9-5 Determine the distance xc to the center of gravity of the homogeneous rod bent into the parabolic shape. If the rod has a weight per unit length determine the reactions at the fixed support O. Given:
= 0.5
a = 1 ft
lb ft
b = 0.5 ft Solution: y=b
x a
2
d y 2bx = 2 dx a L = 0
a
2 b x dx 1+ 2 a
a
2
L = 1.148 ft
1 xc = x L 0
+ Fx = 0;
1+
2 b x dx 2 a
Ox = 0 Ox = 0 lb
2
xc = 0.531 ft
Ox = 0 lb
898
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
+
Fy = 0;
MO = 0;
Oy - L = 0
Oy = L
Oy = 0.574 lb
MO - L xc = 0
MO = L xc
MO = 0.305 lb ft
Problem 9-6 Determine the distance yc to the center of gravity of the homogeneous rod bent into the parabolic shape. Given: a = 1 ft b = 0.5 ft Solution:
x y=b a
2
d y 2bx = 2 dx a L = 0
a
1+
2 b x dx 2 a
2
L = 1.148 ft
a 2 2 1 x 2 b x dx yc = b 1+ 2 L a a 0
yc = 0.183 ft
Problem 9-7 Locate the centroid of the parabolic area. Solution: a= h b
2
899
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
dA = x dy xc = x 2
yc = y
1 h
A= b
0
y y dy = h b h h
h
2
3 xc = 2h b
1 b 2
y dy = h
2
3 8h
2
bh
2
0
xc =
3 b 8
5
3 yc = yb 2h b
0
h
y 3 h dy = h 5
h h
2
yc =
3 h 5
Problem 9-8 Locate the centroid yc of the shaded area. Given: a = 100 mm b = 100 mm
Solution: y = b
x a
2
900
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
x=a
y b y2a
0 b
y dy b yc = 60 mm y b dy
yc =
2a 0
b
Problem 9-9 Locate the centroid xc of the shaded area. Solution: dA = ydx xc = x yc = y 2 h 2 x x dx 2 b
0 b
xc =
2 x h dx 2 b 0
b
b
3 b = 4 3 b
4
xc =
3 b 4
yc =
1 h 2 x dx 2 b2
2
0
b
h 2 x dx 2 b
3 5 h = b 5 10 b h
2
yc =
3 h 10
0
901
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-10 Determine the location (xc, yc) of the centroid of the triangular area.
Solution:
a
1 2 A = m x dx = a m 0 2 xc = 2 2 x m x dx = a 2 3 0 ma
a a
xc =
2 a 3
1 1 2 yc = ( m x) dx = a m 2 2 3 ma 2
0
yc =
m a 3
Problem 9-11 Determine the location (xc, yc) of the center of gravity of the quartercircular plate. Also determine the force in each of the supporting wires.The plate has a weight per unit area of . Given:
= 5
lb ft
2
a = 4 ft
902
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: x +y =a y=
2 2 2 2
a -x
2
2
a A = 4 1 xc = A
W = A
a
W = 62.832 lb
x a2 - x2 dx 0
a
xc = 1.698 ft
1 yc = A Guesses
1 2 2 a - x dx 2 TB = 1lb
(
)
yc = 1.698 ft
0
TA = 1lb
Given
TA + TB - W = 0
TB a - W xc = 0
TA = Find ( TA , TB) TB
TA 36.2 = lb TB 26.7
Problem *9-12 Locate the centroid of the shaded area.
903
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: dA = ydx
L
xc = x
yc =
y 2
2 x A = a sin dx = L a L
0
xc =
2L a
L
x a sin
L
x dx = 1 L 2 L
2
xc =
1 L 2
0
yc = 2L a
1 1 x a sin dx = a 2 8 L
yc =
1 a 8
0
Problem 9-13 Locate the center of gravity of the homogeneous cantilever beam and determine the reactions at the fixed support.The material has a density of . Units Used: Mg = 10 kg Mg m g = 9.81 Solution: V =
0 3 3
kN = 10 N
3
Given:
= 8
a = 1m b = 4m
m s
2
c = 0.5 m
ca
-b 0
x dx b x dx b
2 2
2
W = g V
1 xc = V
xca
-b 0
1 yc = V
-b
c x c a dx 2 b
904
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
1 zc = V Guesses Given
0
-b
-c x a 2 b
2
2
dx
xc -3.00 yc = 0.25 m z -0.30 c
Az = 1 N MA = 1 N m MA - W( b + xc) = 0
Ax = 1 N Ax = 0
Ay = 1 N Ay = 0
Az - W = 0
Ax Ay = Find ( A , A , A , M ) x y z A Az MA
Ax 0.00 Ay = 0.00 kN A 52.32 z
MA = 52.32 kN m
Problem 9-14 Locate the centroid (xc, yc) of the exparabolic segment of area.
Solution: 1 b 2 A= x dx = ab 2 3 a - a 3 -3 b 2 x xc = x dx = a 2 ab 4 a - a 2 3 1 b 2 -3 yc = b - x dx = ab 2 2 10 a - a
0 0 0
xc =
-3 a 4
yc =
-3 b 10
905
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-15 Locate the centroid of the shaded area.
Solution: h n x h - x dx n a
0 a
xc =
a
xc =
h - h xn dx n a
n+1 a 2( n + 2)
provided that n -2
0
yc =
1 2
a 2 h - h xn dx n a
0 a
yc =
h - h xn dx n a
n h 2n + 1
provided that n
-1 2
0
Problem *9-16 Locate the centroid of the shaded area bounded by the parabola and the line y = a.
906
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
3
A= 0
a
2 a a y dy = 3 a
a
( 2)
2
2a A= 3
2
3 xc = 2 2a
1 3 a y dy = a 2 8
5
xc =
3 a 8
0 a
yc =
2 3 3 2 y a y dy = a 2 4 2a 0 5a
( )
yc =
3 a 5
Problem 9-17 Locate the centroid of the quarter elliptical area.
907
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: 2 x dx A = b 1- a
0 a
A=
a b
4
2 4 4 x xc = x b 1 - dx = a a b 3 a
0
a
xc =
4 3
a
2 2 4 1 x b 1 - dx = 4 b yc = a b 2 3 a
0
a
yc =
4 3
b
Problem 9-18 Locate the centroid xc of the triangular area.
Solution: bh 2
a
A=
2 x h x dx + xc = b h a
0
x h ( b - x) dx b-a
b a
xc =
a+b 3
Problem 9-19 Locate the centroid of the shaded area.
908
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 8m b = 4m Solution: A =
a
x 2 b - b1 - dx a
a
A = 10.667 m
2
0
x 2 1 xc = xb - b1 - dx A a
0
xc = 6 m
1 yc = A
a
1 x b + b1 - 2 a
2
x 2 b - b1 - dx a
yc = 2.8 m
0
Problem 9-20 Locate the centroid xc of the shaded area. Solve the problem by evaluating the integrals using Simpson's rule. Given: a = 2 ft
1 5
b = Solution: A = 0
a
a
2
+ 2a
3
1 5 2 3 b - x + 2x dx a
A = 2.177 ft
2
1 xc = x b - A 0
1
5
x
2
+ 2x
3
dx
xc = 0.649 ft
909
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-21 Locate the centroid yc of the shaded area. Solve the problem by evaluating the integrals using Simpson's rule. Given: a = 2 ft
1 5
b =
a
2
+ 2a
3
Solution: A = 0
a 1 5 2 3 b - x + 2x dx
A = 2.177 ft
2
1 yc = A
a
1 5 1 5 1 2 3 2 3 b + x + 2x b - x + 2x dx
2
yc = 2.04 ft
0
Problem 9-22 The steel plate has thickness t and density . Determine the location of its center of mass. Also compute the reactions at the pin and roller support. Units Used: kN = 10 N Given: t = 0.3 m a = 2m kg m
3 3
= 7850
b = 2m c = 2m
910
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
g = 9.81
m s
2
b = atan
c
Solution: A =
c
x b x + a dx c c
c
A = 4.667 m
2
0
1 xc = x a A
0
x b x + dx c c
xc = 1.257 m
1 1 yc = a A 2
0
c
x b x + a c c
x b x - dx c c
yc = 0.143 m
Equilibrium Guesses Given
W = Atg Ax = 1 N Ay = 1 N NB = 1 N - Ax + NB sin ( ) = 0 NB b + c - W xc = 0
2 2
Ay - W + NB cos ( ) = 0
Ax Ay = Find ( Ax , Ay , NB) N B
Ax 33.9 Ay = 73.9 kN N 47.9 B
Problem 9-23 Locate the centroid xc of the shaded area. Given: a = 4 ft b = 4 ft
911
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: A =
a 2 b x x - b dx a a a
0
2 1 b x x dx xc = x - b A a a
0
xc = 2.00 ft
Problem 9-24 Locate the centroid yc of the shaded area. Given: a = 4 ft b = 4 ft
Solution: A =
a 2 b x x dx - b a a
0 a
1 yc = A
1 x x b + b 2 a a
2
2 bx x - b dx a a
yc = 1.60 ft
0
Problem 9-25 Locate the centroid xc of the shaded area. Given: a = 4m
912
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
b = 4m
Solution: A =
a 2 x x b - b dx a a a 2
0
1 xc = xb A
0
x x - b a a
dx
xc = 1.80 m
Problem 9-26 Locate the centroid yc of the shaded area. Given: a = 4m b = 4m
Solution: A =
a 2 x x dx b - b a a a
0
1 yc = A
1 b 2
x x + b a a
2
2 x x dx b - b a a
yc = 1.80 m
0
913
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-27 Locate the centroid xc of the shaded area. Given: a = 1 in b = 3 in c = 2 in
Solution:
A =
a+ b
c
x dx a+b
a
1 xc = A
a+ b
xc
x dx a+b
xc = 2.66 in
a
Problem 9-28 Locate the centroid yc of the shaded area. Given: a = 1 in b = 3 in c = 2 in
914
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: A =
a+ b
c
x dx a+b
a a+ b
1 yc = A
1 c 2
x dx a + b
2
yc = 0.804 in
a
Problem 9-29 Locate the centroid xc of the shaded area. Given: a = 4 in b = 2 in c = 3 in
Solution: A =
a+ b
bc dy y
a+ b
A = 6.592 in
2
b
1 xc = A
1 b c dy 2 y
2
xc = 0.910 in
b
Problem 9-30 Locate the centroid yc of the shaded area.
915
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 4 in b = 2 in c = 3 in Solution: A =
a+ b
bc dy y
A = 6.592 in
2
b a+ b
1 yc = A
y
b c dy y
yc = 3.64 in
b
Problem 9-31 Determine the location rc of the centroid C of the cardioid, r = a(1 - cos).
Solution: A= 0
2
0
a ( 1 - cos( ) )
r dr d =
3 2 a 2
916
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
xc = 2 3a 0 2
2
0
a ( 1 - cos( ) )
r cos ( ) r dr d =
-5 a 6
rc =
5a 6
Problem 9-32 Locate the centroid of the ellipsoid of revolution.
Solution:
dV = z dy
2
z =a
2
2
1-
y b
2 2
2 V = a 1 - 0
b
b
2 2 d y = 1 b 3 b - b a2 2 2 3 b b
y
2
3 2 yc = y a 1 - 2 2b a 0
d y = 3 b2 2 8b b
y By symmetry
2
yc =
3b 8
xc = zc = 0
Problem 9-33 Locate the centroid zc of the very thin conical shell. Hint: Use thin ring elements having a center at (0, 0, z), radius y, and width dL = ( dy) + ( dz)
2 2
917
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
dy a dL = dy + dz = 1 + dz = 1 + dz dz h
2 2
2
2
r=
az h
az A = 2 h
0
h
1+
2 2 2 a dz = h2 a h + a 2 h h h h
1 az zc = z2 h 2 2 a a + h 0
1+
2 h2 + a2 2 2 a dz = h a 2 h 2 2 h h +a 3 a
(
)
zc =
2h 3
Problem 9-34 Locate the centroid zc of the volume.
918
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 2 ft b = 2 ft
Solution: 2 a z V = dz b
0 b
V = 12.566 ft
3
2 1 a z zc = z dz V b
0
b
zc = 1.333 ft
Problem 9-35 Locate the centroid of the solid. Solution: h y a
2
z =
2
y = a
z h
2
xc = yc = 0
h
By symmetry
zc =
2 z 2 z a dz h
0
2 z 2 a dz h 0
h
=
5 h 6 5 h
6
zc =
5h 6
919
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-36 Locate the centroid of the quarter-cone.
Solution: r= a ( h - z) h
h
zc = z
xc = yc =
4r 3
V=
a
4 h
( h - z) dz =
2
1 2 ha 12
0 h
2 12 a dz = 1 h zc = z ( h - z) 2 4 h 4 ha
0
h 2 12 4 a a ( h - z) dz = a xc = ( h - z) 2 h 4 h h a 3 0
xc = yc = a zc = h 4
920
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-37 Locate the center of mass xc of the hemisphere. The density of the material varies linearly from zero at the origin O to o at the surface. Hint: Choose a hemispherical shell element for integration Solution: for a spherical shell x 2
xc =
x = 0
a
2
dV = 2 x dx
xc =
x x 2 x2 dx 0 a 2
0
a
x 2 x2 dx 0 a 0
a
=
2 a 5
xc =
2 a 5
Problem 9-38 Locate the centroid zc of the right-elliptical cone. Given: a = 3 ft b = 4 ft c = 10 ft
x + y = 1 b a
2
2
921
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: Volume and Moment Arm : From the geometry, x b = c-z c y a = c-z c b ( c - z) c a ( c - z) c
x=
y=
The volume of the thin disk differential element is dV = b a ( c - z) ( c - z) dz c c
c
zc =
z b ( c - z) a ( c - z) dz c c
0
b ( c - z) a ( c - z) dz c c 0
c
zc = 2.5 ft
Problem 9-39 Locate the center of gravity zc of the frustum of the paraboloid.The material is homogeneous.
Given: a = 1m b = 0.5 m c = 0.3 m Solution 2 z 2 2 V = b - b - c dz a
a
(
)
0
922
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
1 z b2 - z b2 - c2 dz zc = V a
a
(
)
0
zc = 0.422 m
Problem 9-40 Locate the center of gravity yc of the volume. The material is homogeneous.
Given: a = 25 mm c = 50 mm d = 50 mm Solution:
V =
c+ d
y 2 a d y c
c+ d
2
c
1 yc = V
y 2 y a d y c
2
yc = 84.7 mm
c
Problem 9-41 Locate the center of gravity for the homogeneous half-cone.
923
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: V=
h 2
a y 1 2 dy = h a
2
h
h
6
0
2 6 a y 3 yc = y 2 h dy = 4 h 2 ha
0
yc =
3 h 4
zc = 2 ha 0 6
h 2 4a y a y d y = 1 a 3h 2 h
zc =
a
6 xc = 02 2 ha
0
h
a y dy h
2
xc = 0
Problem 9-42 Locate the centroid zc of the spherical segment.
924
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: 5 3 2 2 V = a - z dz = a 24 a
2 a
(
)
24 z a2 - z2 dz = 27 a zc = 3 40 5 a a
2
a
(
)
zc =
27 a 40
Problem 9-43 Determine the location zc of the centroid for the tetrahedron. Suggestion: Use a triangular "plate" element parallel to the x-y plane and of thickness dz.
925
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: c-z x y = = c a b ab 2 z ( c - z) dz 2 c
0 c
x=
a ( c - z) c
y=
b ( c - z) c
zc =
ab 2 ( c - z) dz c2
0
c
=
1 c 4
zc =
1 c 4
Problem 9-44 Determine the location (x, y) of the particle M1 so that the three particles, which lie in the xy plane, have a center of mass located at the origin O. Given: M1 = 7 kg M2 = 3 kg M3 = 5 kg a = 2m b = 3m c = 4m
Solution: Guesses x = 1m y = 1m M1 y - M2 a - M3 a = 0
Given
M1 x + M2 b - M3 c = 0
x = Find ( x , y) y
x 1.57 = m y 2.29
926
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-45 Locate the center of gravity (xc, yc, zc) of the four particles. Given: M1 = 2 lb M2 = 3 lb M3 = 1 lb M4 = 1 lb f = 4 ft h = -2 ft a = 2 ft b = 3 ft c = -1 ft d = 1 ft e = 4 ft g = 2 ft i = 2 ft Solution:
xc =
M1 0ft + M2 a + M3 d + M4 g M1 + M2 + M3 + M4 M1 0ft + M2 b + M3 e + M4 h M1 + M2 + M3 + M4 M1 0ft + M2 c + M3 f + M4 i M1 + M2 + M3 + M4
xc = 1.29 ft
yc =
yc = 1.57 ft
zc =
zc = 0.429 ft
Problem 9-46 A rack is made from roll-formed sheet steel and has the cross section shown. Determine the location (xc, yc) of the centroid of the cross section. The dimensions are indicated at the center thickness of each segment. Given: a = 15 mm c = 80 mm
927
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 50 mm e = 30 mm
Solution: L = 3a + 2c + e
2a xc =
a 2
+ a e +
a
e + c( a + e) + ea + + ( c - d)a 2 2
L d+c 2 + ad + ec
xc = 24.4 mm
d yc =
d 2
+c
c 2
+ ( c - d) L
yc = 40.6 mm
Problem 9-47 The steel and aluminum plate assembly is bolted together and fastened to the wall. Each plate has a constant width w in the z direction and thickness t. If the density of A and B is s, and the density of C is al, determine the location xc, the center of mass. Neglect the size of the bolts. Units Used: Mg = 10 kg Given: w = 200 mm t = 20 mm a = 300 mm b = 100 mm Mg m
3 3
s = 7.85 al = 2.71
c = 200 mm
Mg m
3
928
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: 2( s a t w) xc = a 2 + al( + b c)t w a - b + b + c 2
2 s a t w + al( b + c)t w
xc = 179 mm
Problem 9-48 The truss is made from five members, each having a length L and a mass density . If the mass of the gusset plates at the joints and the thickness of the members can be neglected, determine the distance d to where the hoisting cable must be attached, so that the truss does not tip (rotate) when it is lifted. Given: L = 4m
= 7
kg m
Solution:
L
d =
L
2
+
L 4
+
3L 4
+L+
5L 4
5 L
d=3m
Problem 9-49 Locate the center of gravity (xc, yc, zc) of the homogeneous wire.
929
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 300 mm b = 400 mm Solution: L =
a
2
+2 a +b
2
2
xc =
1 2 a 2a 2a a + b + L 2 2 1 2 a 2a 2a a + b + L 2 2 1 2 2 b 2 a + b L 2
yc =
zc =
xc 112.2 yc = 112.2 mm z 135.9 c
Problem 9-50 Determine the location (xc, yc) of the center of gravity of the homogeneous wire bent in the form of a triangle. Neglect any slight bends at the corners. If the wire is suspended using a thread T attached to it at C, determine the angle of tilt AB makes with the horizontal when the wire is in equilibrium. Given: a = 5 in b = 9 in c = 12 in
930
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: L = a+b+ a +c +
2 2
b +c
2
2
xc =
1 a+b + ( a + b) L 2 1 2 2c a +c + L 2
a +c
2
2a
2
+
b + c a +
2
2
b 2
xc = 6.50 in
yc =
b +c
2
2 c
2
yc = 4.00 in
= atan
xc - a c - yc
= 10.6 deg
Problem 9-51 The three members of the frame each have weight density . Locate the position (xc,yc) of the center of gravity. Neglect the size of the pins at the joints and the thickness of the members. Also, calculate the reactions at the fixed support A. Given:
= 4
lb ft
P = 60 lb a = 4 ft b = 3 ft c = 3 ft d = 3 ft
931
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: W = d + ( b + c) + 2 d + c + ( a + b)
2 2 2 2
W = 88.774 lb
d + ( b + c)
xc =
2
2d
2 W
+ 2 d + c d xc = 1.6 ft
2
2
( a + b)
yc =
2 2 + d + ( b + c) a + 2
a + b
b + c
2 2 + 2 d + c ( a + b + c) 2
W yc = 7.043 ft
Equilibrium Ax = 0 Ax = 0lb Ay = W + P MA = W xc + P2d Ax = 0 lb Ay = 148.8 lb MA = 502 lb ft
Ay - W - P = 0
MA - W xc - P2d = 0
Problem 9-52 Locate the center of gravity G(xc, yc) of the streetlight. Neglect the thickness of each segment. The mass per unit length of each segment is given. Given: a = 1m b = 3m c = 4m d = 1m e = 1m
AB = 12 BC = 8 CD = 5 DE = 2
f = 1.5 m
kg m kg m kg m kg m
932
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: M = AB c + BC b + CD a + e +
d
2
+ DE f
xc =
1 d 2d f e CD d - + CD ed + + DE fd + e + M 2 2 2 1 a d 2d CDa c + b + 2 + 2 c + b + a + + e( c + b + a + d) ... M b c + DE f( c + b + a + d) + BC bc + + AB c 2 2
yc =
xc 0.200 = m yc 4.365
Problem 9-53 Determine the location yc of the centroid of the beam's cross-sectional area. Neglect the size of the corner welds at A and B for the calculation. Given: d1 = 50 mm d2 = 35 mm h = 110 mm t = 15 mm
Solution:
2 2 d1 d1 d + h + d2 d + h + d2 + h t 1 1 2 2 2 2 2 yc = 2 2 d1 d2 + ht + 2 2
yc = 85.9 mm
933
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-54 The gravity wall is made of concrete. Determine the location (xc, yc) of the center of gravity G for the wall. Given: a = 0.6 m b = 2.4 m c = 0.6 m d = 0.4 m e = 3m f = 1.2 m
Solution:
A = ( a + b + c)d + ( b + c)e -
ce e - ( b + c - f) 2 2
A = 6.84 m
2
xc =
1 2c a + b + c b + c c e ( a + b + c)d 2 + ( b + c)e a + 2 - 2 a + b + 3 ... A b + c - f e + -( b + c - f) a + 3 2 1 d e ce e ( a + b + c)d 2 + ( b + c)e d + 2 - 2 d + 3 ... A e 2e + -( b + c - f) d + 2 3
yc =
xc 2.221 = m yc 1.411
Problem 9-55 Locate the centroid (xc, yc)of the shaded area.
934
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 1 in b = 3 in c = 1 in d = 1 in e = 1 in Solution: A = ( a + b) ( a + e) -
a
4
2
-
1 ( a + b - d) ( a + e - c) 2
2
1 ( a + b) a xc = ( a + e) - A 2 4
2
4a - 1 ( a + b - d) ( a + e - c) a + b - a + b - d 2 3 3
yc =
1 ( a + e) ( a + b) A 2
2
-
a 4a 1 a + e - c - 2 ( a + b - d) ( a + e - c) a + e - 4 3 3
2
xc 1.954 = in yc 0.904
Problem 9-56 Locate the centroid (xc, yc) of the shaded area. Given: a = 1 in b = 6 in c = 3 in d = 3 in Solution: A = bd +
d
4
2
-
a
2
2
+
1 ( d c) 2
935
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
xc =
1 b d b d - A 2 4
2
4d + 1 d c b + 2 3
2
c 3
xc = 2.732 in
yc =
1 d d b d + A 2 4
2 4d - a 4a + 1 d c d 2 3 2 3 3
yc = 1.423 in
Problem 9-57 Determine the location yc of the centroidal axis xcxc of the beam's cross-sectional area. Neglect the size of the corner welds at A and B for the calculation. Given: r = 50 mm t = 15 mm a = 150 mm b = 15 mm c = 150 mm
Solution: b c yc =
+ a t b + 2
b
a
2
2 + r ( b + a + r) 2
b c + a t + r
yc = 154.443 mm
Problem 9-58 Determine the location (xc, yc) of the centroid C of the area. Given: a = 6 in b = 6 in
936
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
c = 3 in d = 6 in
Solution: a b xc = 1 c 1 2 + a c b + + ( b + c)d ( b + c) 3 2 2 3 2 ab + a b yc = a 1 2 ca + 1 2 ( b + c)d b
xc = 4.625 in
1 a 1 d + a c - ( b + c)d 2 2 3 2 3 ab + 1 2 ca + 1 2 ( b + c)d
yc = 1 in
Problem 9-59 Determine the location yc of the centroid C for a beam having the cross-sectional area shown. The beam is symmetric with respect to the y axis. Given: a = 2 in b = 1 in c = 2 in d = 1 in e = 3 in f = 1 in Solution: A = 2[ ( a + b + c + d) ( e + f) - b f - d e]
937
A = 40 in
2
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
yc =
2 ( e + f) ( a + b + c + d) A 2
2
-b
f e - d e f + 2 2
2
yc = 2.00 in
Problem 9-60 The wooden table is made from a square board having weight W. Each of the legs has wieght Wleg and length L. Determine how high its center of gravity is from the floor. Also, what is the angle, measured from the horizontal, through which its top surface can be tilted on two of its legs before it begins to overturn? Neglect the thickness of each leg. Given: W = 15 lb Wleg = 2 lb L = 3 ft a = 4 ft
Solution: W L + 4Wleg zc =
2
L
W + 4Wleg
zc = 2.478 ft
a 2 = atan zc
= 38.9 deg
Problem 9-61 Locate the centroid yc for the beam's cross-sectional area. Given: a = 120 mm
938
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
b = 240 mm c = 120 mm
Solution: A = ( a + b)5c - 3b c
yc =
1 ( a + b) b b 5c - 2b c - b c A 2 2 3
2
yc = 229 mm
Problem 9-62 Determine the location xc of the centroid C of the shaded area which is part of a circle having a radius r.
Solution: A = r - r sin ( ) cos ( )
2 2
xc =
1 2 2r sin ( ) 2 2 - r sin ( ) cos ( ) r cos ( ) r A 3 3
939
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
xc = 2r
sin ( ) 1 - cos ( ) 3 - sin ( ) cos ( )
2
2r xc = 3
sin ( )
3
-
sin ( 2 ) 2
Problem 9-63 Locate the centroid yc for the strut's cross-sectional area. Given: a = 40 mm b = 120 mm c = 60 mm Solution: A =
b
2
2
- 2a c
1 b yc = A 2
2
4b - 2a c c 2 3
yc = 56.6 mm
Problem 9-64 The "New Jersey" concrete barrier is commonly used during highway construction. Determine the location yc of its centroid.
Given: a = 4 in b = 12 in c = 6 in
940
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 24 in
1 = 75 deg 2 = 55 deg
Solution: e = b cot ( 2 ) f = d - 2e h = 1 ( f - a)h 2 2b h 1 2h - 2h e c + b + - ( f - a)hc + b + 3 2 2 3 f-a tan ( 1 ) 2
A = d( c + b + h) - b e - 2h e -
1 ( c + b + h) yc = d A 2 yc = 8.69 in
2
- b e c +
Problem 9-65 The composite plate is made from both steel (A) and brass (B) segments. Determine the mass and location (xc, yc, zc) of its mass center G. Units Used: Mg = 1000 kg Given:
st = 7.85 br = 8.74
Mg m
3
a = 150 mm b = 30 mm c = 225 mm d = 150 mm
Mg m
3
Solution: M = st d b c +
1 1 a b c + br a b c 2 2
xc =
1 1 d 1 a 2a std c b + a b cd + + br a b c d + M 2 2 2 3 3
941
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
yc =
-1 1 b 1 b b st d c b + a b c + br a b c M 2 2 2 2 2 1 1 c 1 2c c std c b + a b c + br a b c M 2 2 2 3 3
zc =
xc 152.8 yc = -15.0 mm z 111.5 c
M = 16.347 kg
Problem 9-66 Locate the centroid yc of the concrete beam having the tapered cross section shown. Given: a = 100 mm b = 360 mm c = 80 mm d = 300 mm e = 300 mm
Solution: ( d + 2e)c yc = 1 b b + ( d - a)bc + + a b c + 2 2 3 2 ( d + 2e)c + 1 2 ( d - a)b + a b c
yc = 135 mm
Problem 9-67 The anatomical center of gravity G of a person can be determined by using a scale and a rigid board having a uniform weight W1 and length l. With the person's weight W known, the person lies down on the board and the scale reading P is recorded. From this show how to calculate the location xc of the center of mass. Discuss the best place l1 for the smooth support at B in order to improve the accuracy of this experiment.
942
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 120 mm b = 240 mm c = 120 mm
Solution: MB = 0; W xc - P l1 + W1 l1 -
l =0 2
P l1 - W 1 l1 - xc =
2
l
W l
Put B as close as possible to the center of gravity of the board, i.e., l1 =
2 the effect of the board's weight will not be a large factor in the measurement.
, then W1 l1 -
2
l
= 0 and
Problem 9-68 The tank and compressor have a mass MT and mass center at GT and the motor has a mass MM and a mass center at GM. Determine the angle of tilt, , of the tank so that the unit will be on the verge of tipping over. Given: a = 300 mm b = 200 mm c = 350 mm d = 275 mm MT = 15 kg MM = 70 kg
943
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: xc = b MT + ( a + b)MM MT + MM c MT + ( c + d)MM MT + MM xc = 0.4471 m
yc =
yc = 0.57647 m
= atan
xc yc
= 37.8 deg
Problem 9-69 Determine the distance h to which a hole of diameter d must be bored into the base of the cone so that the center of mass of the resulting shape is located at zc. The material has a density . Given: d = 100 mm zc = 115 mm
= 8
mg m
3
a = 150 mm b = 500 mm Solution: Guess h = 200 mm
1 Given zc = 3
d h 2 b a b - h
2
4
2 2
2
h = Find ( h)
h = 323 mm
d 1 2 a b - h 3 2
944
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-70 Determine the distance to the centroid of the shape which consists of a cone with a hole of height h bored into its base.
Given: d = 100 mm h = 50 mm
= 8
mg m
3
a = 150 mm b = 500 mm
Solution:
d b h a b - h 3 4 2 2 zc = 2 d 1 2 a b - h 3 2 1
2
2
zc = 128.4 mm
Problem 9-71 The sheet metal part has the dimensions shown. Determine the location (xc, yc, zc) of its centroid. Given: a = 3 in
945
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
b = 4 in c = 6 in Solution: - a b xc = ab +
2
1 2
b xc = -1.143 in
ac
a b yc =
1 2a + a c 2 2 3 ab + 1 2 ac
a
yc = 1.714 in
-1 zc = 2
a c
3
1 2
c zc = -0.857 in
ab +
ac
Problem 9-72 The sheet metal part has a weight per unit area of and is supported by the smooth rod and at C. If the cord is cut, the part will rotate about the y axis until it reaches equilibrium. Determine the equilibrium angle of tilt, measured downward from the negative x axis, that AD makes with the -x axis. Given: a = 3 in b = 4 in c = 6 in
946
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: a b xc = ab +
2
1 2
b xc = 1.143 in
ac
1 zc = 2
a c
3
1 2
c zc = 0.857 in
ab +
ac
= atan
xc zc
= 53.13 deg
Problem 9-73 A toy skyrocket consists of a solid conical top of density t, a hollow cylinder of density c, and a stick having a circular cross section of density s. Determine the length of the stick, x, so that the center of gravity G of the skyrocket is located along line aa. Given: a = 3 mm b = 10 mm c = 5 mm d = 100 mm e = 20 mm Solution: Guess x = 200 mm
2 a2 b e 2 e d x 2 t d + + c b - c d + s x d - = 0
t = 600 c = 400 s = 300
kg m
3
kg m
3
kg m
3
Given
2 3
4
4
(
)
2
4
2
x = Find ( x)
x = 490 mm
947
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-74 Determine the location (xc, yc) of the center of mass of the turbine and compressor assembly. The mass and the center of mass of each of the various components are indicated below. Given: a = 0.75 m b = 1.25 m c = 0.5 m d = 0.75 m e = 0.85 m f = 1.30 m g = 0.95 m M1 = 25 kg M2 = 80 kg M3 = 30 kg M4 = 105 kg
Solution: M = M1 + M2 + M3 + M4 1 M2 a + M3 ( a + b) + M4( a + b + c) M 1 ( M 1 d + M 2 e + M 3 f + M 4 g) M
xc =
xc = 1.594 m
yc =
yc = 0.940 m
Problem 9-75 The solid is formed by boring a conical hole into the hemisphere. Determine the distance zc to the center of gravity.
948
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: V=
3 2 3 2 a - a a = a 3 3 3
1 5a 2 3 3 3 a - a a V 8 3 4 3
zc =
zc =
a 2
Problem 9-76 Determine the location xc of the centroid of the solid made from a hemisphere, cylinder, and cone. Given: a = 80 mm b = 60 mm c = 30 mm d = 30 mm Solution: V = 1 2 2 3 2 d a + d b + d 3 3 1 1 2 3a b 2 3 3c 2 d a + d b a + + d a + b + V 3 8 4 2 3
xc =
xc = 105.2 mm
949
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-77 The buoy is made from two homogeneous cones each having radius r. Find the distance zc to the buoy's center of gravity G. Given: r = 1.5 ft h = 1.2 ft a = 4 ft
Solution:
2 a 2 h r a - r h
zc = 3
4
2
3
3
4
zc = 0.7 ft
r ( a + h)
Problem 9-78 The buoy is made from two homogeneous cones each having radius r. If it is required that the buoy's center of gravity G be located at zc,determine the height h of the top cone. Given: zc = 0.5 ft r = 1.5 ft a = 4 ft
Solution: Guess h = 1 ft
950
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
2 a 2 h r a - r h
Given zc = 3
4
2
3
3
4
h = Find ( h)
h = 2 ft
r ( a + h)
Problem 9-79 Locate the center of mass zc of the forked lever, which is made from a homogeneous material and has the dimensions shown. Given: a = 0.5 in b = 2.5 in c = 2 in d = 3 in e = 0.5 in
Solution:
V = b a + 2e a d +
2
2
2 2 ( c + e) - c a
zc =
1 2 b d a 2 c + e b a 2 + 2e a db + e + c + 2 + 2 ( c + e) b + c + e - 4 3 ... V - a c2 b + c + e - 4 c + 2 3
zc = 4.32 in
951
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-80 A triangular plate made of homogeneous material has a constant thickness which is very small. If it is folded over as shown, determine the location yc of the plate's center of gravity G. Given: a = 6 in b = 3 in c = 1 in d = 3 in e = 1 in f = 3 in
Solution:
2d b yc =
1 2b 1 f + ( 2c b) + ( 2e f) 2 2 3 2 3 2d b + 1 2 ( 2c b) + 1 2 ( 2d) ( a + f)
b
yc = 0.75 in
Problem 9-81 A triangular plate made of homogeneous material has a constant thickness which is very small. If it is folded over as shown, determine the location zc of the plate's center of gravity G. Given a = 6 in b = 3 in c = 1 in
952
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 3 in e = 1 in f = 3 in
Solution:
1 zc = 2
( 2e f a) + 2e a 2d b + 1 2
1 a + 2( d - e)a 2 2 3 1 2 2d( a + f)
a
zc = 1.625 in
( 2c b) +
Problem 9-82 Each of the three homogeneous plates welded to the rod has a density and a thickness a. Determine the length l of plate C and the angle of placement, , so that the center of mass of the assembly lies on the y axis. Plates A and B lie in the xy and zy planes, respectively. Units Used: Mg = 1000 kg Given: a = 10 mm b = 200 mm c = 250 mm f = 100 mm g = 150 mm e = 150 mm
= 6
Solution: Guesses
Mg m
3
The thickness and density are uniform
= 10 deg
l = 10 mm
953
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given b f
f - g l g cos ( ) = 0 2 2
-c e
e + g l g sin ( ) = 0 2 2
= 70.4 deg
l = Find ( l , )
l = 265 mm
Problem 9-83 The assembly consists of a wooden dowel rod of length L and a tight-fitting steel collar. Determine the distance xc to its center of gravity if the specific weights of the materials are w and st.The radii of the dowel and collar are shown. Given: L = 20 in
w = 150 st = 490
a = 5 in b = 5 in r1 = 1 in r2 = 2 in Solution:
lb ft
3
lb ft
3
w r1 L
xc =
) 2 2 2 2 w r1 L + st ( r2 - r1 ) b
2 L 2 2
+ st r2 - r1 b a +
(
b
2
xc = 8.225 in
Problem 9-84 Determine the surface area and the volume of the ring formed by rotating the square about the vertical axis.
954
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
= 45 deg
Solution:
A = 22 b -
a sin ( ) a ... 2 2 b + a sin ( ) a + 2 2
A = 8 b a V = 2 b a
2
Problem 9-85 The anchor ring is made of steel having specific weight st. Determine the surface area of the ring. The cross section is circular as shown. Given:
st = 490
a = 4 in b = 8 in
lb ft
3
955
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: A = 2
a + b - a 2 b - a 4 4 2
A = 118 in
2
Problem 9-86 Using integration, determine both the area and the distance yc to the centroid of the shaded area. Then using the second theorem of PappusGuldinus, determine the volume of the solid generated by revolving the shaded area about the x axis. Given: a = 1 ft b = 2 ft c = 2 ft
Solution: A =
c
y2 a + b d y c
c
A = 3.333 ft
2
0
2 1 y yc = ya + b d y A c
0
yc = 1.2 ft
V = 2 yc A
V = 25.1 ft
3
956
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-87 The grain bin of the type shown is manufactured by Grain Systems, Inc. Determine the required square footage of the sheet metal needed to form it, and also the maximum storage capacity (volume) within it. Given: a = 30 ft b = 20 ft c = 45 ft
Solution:
A = 2 a c + 2
a 2
a +b
2
2
A = 11.9 10 ft
3 2
V = 2 a c
a + 2 2
3 3
a1 a b 32
V = 146 10 ft
Problem 9-88 Determine the surface area and the volume of the conical solid.
957
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: 3 a 2 2 2
2
A = 2a
A=
3 a
V = 2
1 a 3 a 3 a2 2 2 2 6
V=
3
4 a
Problem 9-89 Sand is piled between two walls as shown. Assume the pile to be a quarter section of a cone and that ratio p of this volume is voids (air space). Use the second theorem of Pappus-Guldinus to determine the volume of sand. Given: r = 3m h = 2m p = 0.26 Solution:
V = ( 1 - p)
r h r 2 3 2
V = 3.487 m
3
958
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-90 The rim of a flywheel has the cross section A-A shown. Determine the volume of material needed for its construction. Given: r = 300 mm a = 20 mm b = 40 mm c = 20 mm d = 60 mm
Solution: V = 2 r + b +
6
c b d c + 2 r + b a 2 2
3
V = 4.25 10 mm
Problem 9-91 The Gates Manufacturing Co. produces pulley wheels such as the one shown. Determine the weight of the wheel if it is made from steel having a specific weight . Given: a = 1 in c = 0.5 in d = 1 in e = 1 in f = 0.25 in b = 2( c + d + e)
= 490
lb ft
3
959
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: W = 2 d a c +
d a a - f + c + d + e 2 3 2
W = 3.01 lb
Problem 9-92 The Gates Manufacturing Co. produces pulley wheels such as the one shown. Determine the total surface area of the wheel in order to estimate the amount of paint needed to protect its surface from rust. Given: a = 1 in c = 0.5 in d = 1 in e = 1 in f = 0.25 in b = 2( c + d + e) Solution: A = 2 f( c + d) + a c + 2( d + e) c +
d + e 2 a - f +2 e + 2 2
2
c + d +
e 2
A = 70 in
2
Problem 9-93 Determine the volume of material needed to make the casting. Given: r1 = 4 in r2 = 6 in
960
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
r3 = r2 - r1 Solution:
V = 2 2
r 2 4r2 + 2r 2r r2 - 2 r 2 r - 4r3 2 3 2 2( 3) 3 4 3 2 2
3 3
V = 1.40 10 in
Problem 9-94 A circular sea wall is made of concrete. Determine the total weight of the wall if the concrete has a specific weight c. Given:
c = 150
a = 60 ft b = 15 ft c = 8 ft d = 30 ft
lb ft
3
= 50 deg
Solution: c 2 1 1 W = c a d( b - c) + ( b - c) d( b - c) + a + b - d c 2 2 3 2 W = 3.12 10 lb
6
Problem 9-95 Determine the surface area of the tank, which consists of a cylinder and hemispherical cap.
961
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 4m b = 8m
Solution: A = 2 a b + A = 302 m
2
2a a 2
Problem 9-96 Determine the volume of the tank, which consists of a cylinder and hemispherical cap. Given: a = 4m b = 8m Solution: V = 2
4a a2 a + ( b a) 3 4 2
3
V = 536 m
962
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-97 Determine the surface area of the silo which consists of a cylinder and hemispherical cap. Neglect the thickness of the plates. Given: a = 10 ft b = 10 ft c = 80 ft Solution: A = 2
2a a + a c 2
3 2
A = 5.65 10 ft
Problem 9-98 Determine the volume of the silo which consists of a cylinder and hemispherical cap. Neglect the thickness of the plates. Given: a = 10 ft b = 10 ft c = 80 ft Solution: V = 2
4a a2 a + c a 2 3 4
3 3
V = 27.2 10 ft
963
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-99 The process tank is used to store liquids during manufacturing. Estimate both the volume of the tank and its surface area. The tank has a flat top and the plates from which the tank is made have negligible thickness. Given: a = 4m b = 6m c = 3m Solution: V = 2
c c a + c ( c b) 3 2 2
3 2
V = 207 m
c c A = 2 c + c b + 2 2 A = 188 m
2
a +c
2
Problem 9-100 Determine the height h to which liquid should be poured into the cup so that it contacts half the surface area on the inside of the cup. Neglect the cup's thickness for the calculation. Given: a = 30 mm b = 50 mm c = 10 mm Solution: Total area Atotal = 2 c Guess h = 1 mm
c + a + c b2 + ( a - c) 2 2 2
e = 1 mm
964
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given a-c e-c = b h Atotal 2 = 2 c
c + e + c h2 + ( e - c) 2 2 2
e = 21.942 mm h = 29.9 mm
e = Find ( e , h) h
Problem 9-101 Using integration, compute both the area and the centroidal distance xc of the shaded region. Then, using the second theorem of PappusGuldinus, compute the volume of the solid generated by revolving the shaded area about the aa axis. Given: a = 8 in b = 8 in Solution:
2 x A = b dx a
0
a
2 1 x dx xc = 2a - x b A a
0
a
A = 21.333 in
2
xc = 10 in
V = 2 A xc
V = 1.34 10 in
3
3
965
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-102 Using integration, determine the area and the centroidal distance yc of the shaded area. Then, using the second theorem of PappusGuldinus, determine the volume of a solid formed by revolving the area about the x axis. Given: a = 0.5 ft b = 2 ft c = 1 ft Solution: A =
b
c dx x
b
2
A = 1.386 ft
2
a
1 yc = A
1 c 2 x
2
2
dx
yc = 0.541 ft
a
V = 2 A yc
V = 4.71 ft
3
Problem 9-103 Determine the surface area of the roof of the structure if it is formed by rotating the parabola about the y axis.
Given: a = 16 m b = 16 m
966
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: Centroid : The length of the differential element is dL = dx + dy =
2 2 2 1 + dy dx dx
and its centroid is
xc = x
Here,
dy bx = -2 2 dx a
Evaluating the integrals, we have L = 0
a 2 2 4
1+
4b x a
dx
L = 23.663 m
2 2 1 4b x dx xc = x 1 + 4 L a 0
a
xc = 9.178 m
A = 2 xc L
A = 1.365 10 m
3
2
Problem 9-104 The suspension bunker is made from plates which are curved to the natural shape which a completely flexible membrane would take if subjected to a full load of coal.This curve may be approximated by a parabola, y/b = (x/a)2. Determine the weight of coal which the bunker would contain when completely filled. Coal has a specific weight of , and assume there is a fraction loss p in volume due to air voids. Solve the problem by integration to determine the cross-sectional area of ABC; then use the second theorem of PappusGuldinus to find the volume. Units Used: kip = 10 lb Given: a = 10 ft b = 20 ft
967
3
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
= 50
lb ft
3
p = 0.2
Solution: A = a
0 b b
y dy b
A = 133.3 ft
2
1 xc = A
1 a 2
y dy b
2
xc = 3.75 ft
0
V = 2 A xc
V = 3.142 10 ft
3 3
W = ( 1 - p) V
W = 125.7 kip
Problem 9-105 Determine the interior surface area of the brake piston. It consists of a full circular part. Its cross section is shown in the figure. Given: a = 40 mm b = 30 mm c = 20 mm d = 20 mm e = 80 mm
968
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
f = 60 mm g = 40 mm Solution: c 3c a b 2 2 A = 2 a + a + b + e + c a + b + + ( a + b + c) f + a + b + c ... 2 2 2 2 + ( a + b + 2c)g A = 119 10 mm
3 2
Problem 9-106 Determine the magnitude of the resultant hydrostatic force acting on the dam and its location H, measured from the top surface of the water. The width of the dam is w; the mass density is w. Units Used: Mg = 10 kg MN = 10 N Given: w = 8m
6 3
w = 1
h = 6m g = 9.81 Solution:
Mg m
3
m s
2
p = h w g F = 1 hw p 2
p = 58860
N m
2
F = 1.41 MN H=4m
H =
2 h 3
969
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-107 The tank is filled with water to a depth d. Determine the resultant force the water exerts on side A and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so that it creates the same resultant forces? The densities are 0 and w. Given: kN = 10 N
3
d = 4m a = 3m b = 2m
o = 900
kg m
3
w = 1000
g = 9.81 Solution: For water At side A: m s
2
kg m
3
WA = b w g d F RA = 1 WA d 2
WA = 78480
N m
F RA = 157 kN N m
At side B:
WB = a w g d
WB = 117720
F RB = For oil At side A:
1 WB d 2
F RB = 235 kN
F RA =
1 b o g d1 d1 2
d1 =
2FRA b o g
d1 = 4.216 m
970
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-108 The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment about O due to the dam's weight divided by the overturning moment about O due to the water pressure. Determine this factor if the concrete has specific weight conc and water has specific weight w. Given: a = 3 ft b = 15 ft c = 9 ft
w = 62.4
lb ft
3
conc = 150
lb ft
3
Solution: For a 1-ft thick section: W = w b( 1ft) F = 1 Wb 2 W = 936 lb ft
F = 7020 lb W1 = 6750 lb
W1 = conc( 1ft)a b 1 W2 = conc ( c - a)b( 1ft ) 2 Moment to overturn: 1 MO = F b 3 Moment to stabilize: MS = W1 ( c - a) + Fs = MS MO
W2 = 6750 lb
MO = 35100 lb ft
a 2 + W2 ( c - a) 2 3
MS = 77625 lb ft F s = 2.21
971
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-109 The concrete "gravity" dam is held in place by its own weight. If the density of concrete is c and water has a density w, determine the smallest dimension d that will prevent the dam from overturning about its end A. Units Used: Mg = 10 kg Given:
3
c = 2.5 w = 1.0
h = 6m g = 9.81 Solution:
Mg m
3
Mg m m s
2 3
Consider a dam of width a = 1 m. w = w g h a W= 1 c g d h a 2 W 2d h -F =0 3 3 w = 58860 N m F = 1 wh 2 F = 176580 N
Equilibrium
1 Fh 2d c g d h a = 2 3 3 F
d =
c g a
d = 2.683 m
Problem 9-110 The concrete dam is designed so that its face AB has a gradual slope into the water as shown. Because of this, the frictional force at the base BD of the dam is increased due to the hydrostatic force of the water acting on the dam. Calculate the hydrostatic force acting on the face AB of the dam. The dam has width w, the water density is w.
972
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Units Used: kip = 10 lb Given: w = 60 ft
3
w = 62.4
a = 18 ft b = 12 ft Solution: F AB =
lb ft
3
1 2 2 w w b a + b 2
F AB = 486 kip
Problem 9-111 The symmetric concrete "gravity" dam is held in place by its own weight. If the density of concrete is c and water has a density w, determine the smallest distance d at its base that will prevent the dam from overturning about its end A.The dam has a width w. Units Used: Mg = 10 kg Given: a = 1.5 m b = 9m w = 8m Solution: Guesses d = 3m F v = 1 MN F h = 1 MN W = 1 MN
3
MN = 10 N
6
c = 2.5 w = 1.0
Mg m
3
Mg m
3
973
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given Fv = Fh = d-a b w w g 4 1 w g b wb 2
W = c g wa b +
d - a b 2
W
d d - a - F b = 0 + F v d - h 3 2 6
Fv Fh = Find ( F , F , W , d) v h W d
Fv 0.379 Fh = 3.178 MN W 4.545
d = 3.65 m
Problem 9-112 The tank is used to store a liquid having a specific weight . If it is filled to the top, determine the magnitude of force the liquid exerts on each of its two sides ABDC and BDFE. Units used: kip = 10 lb Given:
3
= 80
lb ft
3
a = 6 ft b = 6 ft c = 12 ft d = 8 ft e = 4 ft
974
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution: Fluid Pressure: The fluid pressure at points B and E can be determined using pB = e pB = 320 lb ft pE = ( e + d) Thus wB = pB c wE = pE c wB = 3.84 kip ft kip ft pE = 960
2
lb ft
2
wE = 11.52
Resultant Forces: The resultant Force acts on surface ABCD is 1 2 2 wB e + b 2
F R1 =
F R1 = 13.8 kip
and on surface BDFE is F R2 = 1 (wB + wE)d 2 F R2 = 61.4 kip
Problem 9-113 The rectangular gate of width w is pinned at its center A and is prevented from rotating by the block at B. Determine the reactions at these supports due to hydrostatic pressure. Units Used: Mg = 10 kg Given: a = 1.5 m b = 6m
3
kN = 10 N
3
w = 1.0
Mg m
3
975
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
w = 2m Solution:
g = 9.81
m s
2
w1 = w g( b - 2a)w w2 = w g2a w F1 = 1 2a w1 2
w1 = 59 w2 = 59
kN m kN m
F 1 = 88 kN F 2 = 177 kN
F 2 = w2 2a
MA = 0;
F1
a - FB a = 0 3
FB =
1 F1 3
F B = 29.4 kN F A = 235 kN
F x = 0;
F1 + F2 - FB - FA = 0
FA = F1 + F2 - FB
Problem 9-114 The gate AB has width w. Determine the horizontal and vertical components of force acting on the pin at B and the vertical reaction at the smooth support A. The density of water is w. Units Used: Mg = 10 kg kN = 10 N MN = 10 N Given: w = 8m
6 3 3
w = 1.0
a = 5m b = 4m
Mg m
3
976
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
c = 3m g = 9.81 Solution: Fluid Pressure: The fluid pressure at points A and B can be determined using Eq. 9-15, pA = w g( a + b) pA = 88.29 kN m pB = w g a Equilibrium b +c 1 b +c + ( wA - wB) - Ay c = 0 2 2 3 b +c 2
2 2 2 2 2 2 2
m s
2
wA = pA w
wA = 706.32
kN m
pB = 49.05
kN m
2
wB = pB w
wB = 392.4
kN m
wB
wB Ay =
+
(wA - wB) 2
1 c
2b +c 3
(2
2
)
Ay = 2.507 MN
Ay - wB c -
1 (wA - wB)c - By = 0 2 1 (wA - wB)c 2 B y = 858.92 kN
B y = Ay - wB c - -B x + wB b +
1 (wA - wB)b = 0 2 1 (wA - wB)b 2 B x = 2.197 MN
B x = wB b +
Problem 9-115 The storage tank contains oil having a specific weight . If the tank has width w, calculate the resultant force acting on the inclined side BC of the tank, caused by the oil, and specify its location along BC, measured from B. Also compute the total resultant force acting on the bottom of the tank.
977
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Units Used: kip = 10 lb Given:
3
= 56
w = 6 ft
lb ft
3 c = 8 ft
d = 4 ft e = 3 ft f = 4 ft
a = 10 ft b = 2 ft
Solution: wB = w b F h1 = wB c The resultant force F Rx = F h1 + F h2 F Ry = F v1 + Fv2 FR = FRx + FRy Given eh c +e
2 2 2 2
wB = 672
lb lb wC = w ( b + c) wC = 3360 ft ft 1 1 F h2 = ( wC - wB) c F v1 = w b e F v2 = w c e 2 2
F R = 17.225 kip
The location h measured from point B
Guess h = 1 ft ch c +e
2 2
F v1
e 2e c 2c + F v2 + F h1 + Fh2 = F Rx 2 3 2 3
+ F Ry
On the bottom of the tank F bot = w f( b + c + d)
h = Find ( h)
h = 5.221 ft F bot = 18.816 kip
Problem 9-116 The arched surface AB is shaped in the form of a quarter circle. If it has a length L, determine the horizontal and vertical components of the resultant force caused by the water acting on the surface. The density of water is w.
978
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Units Used: Mg = 10 kg kN = 10 N Given: L = 8m
3 3
w = 1.0
a = 3m b = 2m g = 9.81 Solution:
Mg m
3
m s
2
F3 = w g a b L F2 = w g a b L b F1 = w g b L 2 W = b -
2
F 3 = 470.88 kN F 2 = 470.88 kN F 1 = 156.96 kN
2 b
4
L w g
W = 67.368 kN F x = 628 kN F y = 538 kN
Fx = F1 + F2 Fy = F2 + W
Problem 9-117 The rectangular bin is filled with coal, which creates a pressure distribution along wall A that varies as shown, i.e. p = p0(z/b)3. Determine the resultant force created by the coal and specify its location measured from the top surface of the coal. Units used: kip = 10 lb
3
979
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 4 ft b = 10 ft p0 = 4000 lb ft Solution: Resultant Force and its location: F =
b 3 2
p0
z a dz b
F = 40 kip
0 b
3 1 z zc = z p0 a dz F b
0
zc = 8 ft
Problem 9-118 The semicircular drainage pipe is filled with water. Determine the resultant horizontal and vertical force components that the water exerts on the side AB of the pipe per foot of pipe length; water has density .
Given:
= 62.4
r = 2 ft
lb ft
3
Solution: w = r w = 124.8 lb ft Resultant forces (per unit foot): F Rh = 1 wr 2 F Rh = 124.8 lb ft
2
980
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
F Rv =
r
4
2
F Rv = 196.0
lb ft
Problem 9-119 The load over the plate varies linearly along the sides of the plate such that p = k y (a-x). Determine the magnitude of the resultant force and the coordinates (xc, yc) of the point where the line of action of the force intersects the plate. Given: a = 2 ft b = 6 ft k = 10 lb ft
4
Solution: p ( x , y) = k y( a - x) F R = p ( x , y) d y dx 0 0 1 x p ( x , y) d y dx FR 0 0 1 y p ( x , y) d y dx F R 0 0
a b a b a b
F R = 360 lb
xc =
xc = 0.667 ft
yc =
yc = 4 ft
Problem 9-120 The drum is filled to its top (y = a) with oil having a density . Determine the resultant force of the oil pressure acting on the flat end of plate A of the drum and specify its location measured from the top of the drum.
981
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: a = 1.5 ft
= 55
lb ft
3
Solution: 2 2 F R = 2 a - y ( a - y) d y - a 1 y 2 a2 - y2 ( a - y) d y d = a- FR - a
a a
F R = 583 lb
d = 1.875 ft
Problem 9-121 The gasoline tank is constructed with elliptical ends on each side of the tank. Determine the resultant force and its location on these ends if the tank is half full. Given: a = 3 ft b = 4 ft
= 41
lb ft
3
Solution: FR =
0
- y2
-a
b a2 - y2 d y a
F R = 984 lb
982
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
1 yc = FR
0
y- y2
-a
b a2 - y2 d y a
yc = -1.767 ft
xc = 0 ft
Problem 9-122 The loading acting on a square plate is represented by a parabolic pressure distribution. Determine the magnitude of the resultant force and the coordinates (xc, yc) of the point where the line of action of the force intersects the plate. Also, what are the reactions at the rollers B and C and the ball-and-socket joint A? Neglect the weight of the plate. Units Used: kPa = 10 Pa kN = 10 N Given: a = 4m p0 = 4 kPa Solution: Due to symmetry xc = 0 FR =
a 3 3
p0
a
y a dy a
F R = 42.667 kN
0
1 yc = y p0 FR
0
y a dy a
yc = 2.4 m
Equilibrium
Guesses
Ay = 1 kN
B y = 1 kN
Cy = 1 kN
Given
Ay + B y + Cy - F R = 0
(By + Cy)a - FR yc = 0
By a a - Cy = 0 2 2
983
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Ay By = Find ( Ay , By , Cy) C y
Ay 17.067 By = 12.8 kN C 12.8 y
Problem 9-123
The tank is filled with a liquid which has density . Determine the resultant force that it exerts on the elliptical end plate, and the location of the center of pressure, measured from the x axis.
Units Used: kN = 10 N Given: a = 1m b = 0.5 m
3
= 900
kg m
3
g = 9.81 Solution: FR =
b
m s
2
-b
y g2a 1 - ( b - y) d y b
b
2
F R = 6.934 kN
1 yc = FR
y g2a 1 -
-b
y ( b - y) d y b
2
yc = -0.125 m
Problem 9-124 A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the volume of material required to make the belt.
984
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given: r = 600 mm a = 25 mm b = 50 mm c = 75 mm Solution:
V = 2 r +
c 1 c 2 a c + r + b c 3 2 2
V = 22.4 10
-3
m
3
Problem 9-125 A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the surface area of the belt. Given: r = 600 mm a = 25 mm b = 50 mm c = 75 mm Solution:
A = 2 r b + 2 r +
c 2
a + c + ( r + c) ( b + 2a)
2
2
A = 1.246 m
2
985
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-126 Locate the center of mass of the homogeneous rod. Given: a = 200 mm b = 600 mm c = 100 mm d = 200 mm
= 45 deg
Solution: L = a+b+c+d xc = 1 b b sin ( ) + c b sin ( ) L 2 1 d b d + b cos ( ) + c b cos ( ) L 2 2 1 a c a + d a - c L 2 2 xc = 154.3 mm
yc =
yc = 172.5 mm
zc =
zc = 50.0 mm
Problem 9-127 Locate the centroid of the solid
986
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
2a
z a a -
2a
z dz 2 = z
yc =
0
2a -a z
4
3
0
a a -
2
dz
a
3
yc =
2 a 3
Problem 9-128 Locate the centroid (xc, yc) of the thin plate. Given: a = 6 in
Solution: a a - 2 4
2 2 2
A = 4a - 1 -a A 2
2
A = 97.7 in
2
xc =
2 -2 a - a a - 4a 4 3 3 2
xc = -0.262 in
1 -a 2a a yc = - A 2 3 4
2
4a - a 3
yc = 0.262 in
987
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-129 Determine the weight and location (xc, yc) of the center of gravity G of the concrete retaining wall. The wall has a length L, and concrete has a specific gravity of . Units Used: kip = 10 lb Given: a = 12 ft b = 9 ft c = 1.5 ft d = 5.5 ft e = 1.5 ft Solution: A = bc + a f + 1 a( e - f ) 2 W = AL W = 42.8 kip f = 1 ft g = 2 ft L = 10 ft
3
= 150
lb ft
3
xc =
1 b f 1 e - f b c + a fg + + a( e - f) g + f + A 2 3 2 2 1 c a 1 a b c + a f c + + a( e - f) c + A 2 2 2 3
xc = 3.52 ft
yc =
yc = 4.09 ft
Problem 9-130 The hopper is filled to its top with coal. Determine the volume of coal if the voids (air space) are a fraction p of the volume of the hopper. Given: a = 1.5 m b = 4m c = 1.2 m
988
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 0.2 m p = 0.35
Solution:
V = ( 1 - p)2 d c V = 20.5 m
3
d 1 a - d + a b a + c( a - d) d + 2 2 3 2
Problem 9-131 Locate the centroid (xc, yc) of the shaded area.
Given: a = 16 ft b = 4 ft c =
(
a-
b)
2
Solution: A = 0
b
(
a-
x) dx
2
A = 29.3 ft
2
989
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
1 xc = A
x( a - 0
b
b
x) dx
2 2
xc = 1.6 ft
1 yc = A
( a - x) 2 dx
2
yc = 4.15 ft
0
Problem 9-132 The rectangular bin is filled with coal, which creates a pressure distribution along wall A that varies as shown, i.e., p = p0(z/b)1/3. Compute the resultant force created by the coal, and its location, measured from the top surface of the coal. Given: p0 = 8 a = 3 ft b = 8 ft lb ft
2
Solution: F =
b 1
z p0 a dz b
b
3
F = 144 lb
0
1 3 1 z zc = z p0 a dz F b
0
zc = 4.57 ft
990
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-133 The load over the plate varies linearly along the sides of the plate such that p = Determine the resultant force and its position (xc, yc) on the plate. 2 x( 4 - y) kPa 3
Solution: F =
3
0
4
2 x( 4 - y) d y dx 3
4
F = 24 kN
0
1 x 2 x( 4 - y) d y dx xc = F 3
0 0 3 4
3
xc = 2 m
1 y 2 x( 4 - y) d y dx yc = F 3
0 0
yc = 1.333 m
Problem 9-134 The pressure loading on the plate is described by the function p = { -240/(x + 1) + 340 } Pa. Determine the magnitude of the resultant force and coordinates of the point where the line of action of the force intersects the plate. Solution: Due to symmetry yc = 3 m F =
5
-240 + 340 6 dx x + 1
0
991
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
F = 7.62 10 N
5
3
1 -240 x xc = + 340 6 dx F x + 1
0
xc = 2.74 m
992
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.