Complete COSMOS: Online Solutions Manual Organization System
Chapter 16, Problem 1.
A 30-kg uniform thin panel is placed in a truck with end A resting on a rough horizontal surface and end B supported by a smooth vertical surface. Knowing that the deceleration of the truck is 4 m/s2, determine (a) the reactions at ends A and B, (b) the minimum required coefficient of static friction at end A.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 1.
M A = N B (1.5 m ) sin 60 - ( 294.3 N )( 0.75 m )( cos 60 )
= ( 30 kg ) 4 m/s 2 ( sin 60 )( 0.75 m )
N B = 144.96 N
(
)
Fx = N B - F = 30 kg 4 m/s 2 F = 24.96 N
(
)
(a)
RA =
2 N A + F 2 = 295.36 N
or R A = 295 N
85.2 !
= tan - 1
294.3 = 85.2 24.96
and B = 145.0 N (b)
!
=
F 24.96 = = 0.08481 N A 294.3
or = 0.0848 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 3.
Two identical 0.9-lb slender rods AB and BC are welded together to form an L-shaped assembly. The assembly is guided by two small wheels that roll freely in inclined parallel slots cut in a vertical plate. Knowing that = 30, determine (a) the acceleration of the assembly, (b) the reactions at A and C.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 3.
Fx = ma
(a)
Fx = mg ( cos 60 ) = ma
a=
g 2 a = 16.10 ft/s 2 !
M C = ( M C )eff
(b)
M C = - (1.8 lb )( 2 in.) + RA ( cos 30 ) ( 8 in.) - RA ( sin 30 )( 8 in.) -1.8 1.8 = ( a )( cos 30 )( 6 in.) - ( a )( sin 30 )( 2 in.) g g - 3.6 + RA ( 2.9282) = - (11.1531) a g
1 3.6 - 11.1531 2 RA = = - 0.675 lb 2.9282
or R A = 0.675 lb
Fy = RA + RC - (1.8 lb ) cos 30 = 0 RA + RC = 1.5588 RC = 1.5588 - ( - 0.675) = 2.234 lb
60 !
or R C = 2.23 lb
60 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 11.
Bars AB and BE, each of weight 8 lb, are welded together and are pin-connected to two links AC and BD. Knowing that the assembly is released from rest in the position shown and neglecting the masses of the links, determine (a) the acceleration of the assembly, (b) the forces in the links.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 11.
M G = 0 FA = 0
F BG = ( mg )( sin 30 ) = ma
a= g 2 or a = 16.1 ft/s 2
F BG = FB - (16 lb )( cos 30 ) = 0
FB = 13.856 lb
(a) (b)
a = 16.10 ft/s 2
30
FA = 0, FB = 13.86 lb compression
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
(c) Front-wheel drive:
M A = ( M A )eff :
( 2.5 m ) N B - (1.5 m )W
= - ( 0.5 m ) ma
N B = 0.6W - 0.2ma
Thus:
FB = k N B = 0.80 ( 0.6W - 0.2ma ) = 0.48mg - 0.16ma Fx = ( Fx )eff : FB = ma 0.48mg - 0.16ma = ma 0.48 g = 1.16a a = 0.48 9.81 m/s 2 = 4.0593 m/s 2 1.16
(
)
or a = 4.06 m/s 2
!
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 16.
At the instant shown the tensions in the vertical ropes AB and DE are 300 N and 200 N, respectively. Knowing that the mass of the uniform bar BE is 5 kg, determine, at this instant, (a) the force P, (b) the magnitude of the angular velocity of each rope, (c) the angular acceleration of each rope.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 16.
M G = 0 = 200 ( 0.6 )( 0.866 ) - 300 ( 0.6 )( 0.866 ) + P ( 0.6 )( 0.5)
(a)
P = 173.2 N
!
Fy = 300 - 5 ( 9.81) + 200 = 5 ( 0.4 ) 2
(b)
= 15.02 rad/s !
Fx = P = 5 ( 0.4 ) = 173.2
(c)
= 86.6 rad/s 2
!
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 21.
A 200-kg flywheel is at rest when a constant 300 N m couple M is applied at time t = 0. At t = 28 s the flywheel reaches its rated speed of 2400 rpm and the couple is removed. Knowing that the radius of gyration of the flywheel is 400 mm, determine (a) the average magnitude of the couple due to kinetic friction in the bearing, (b) the time at which the flywheel comes to rest.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 21.
M 0 = I = mk 2 ,
300 - M F = 200 ( 0.4 ) = 32
2
MF from friction
(a) Kinematics
1 = t1,
t1 = 28 s
1 = 2400
2 = 80 rad/s 60
2 80 300 - M F = 200 ( 0.4 ) = 287.23 28
M F = 12.77 N m
!
(b) I 2 = -12.77 N m, Kinematics
2 =
T
-12.7687 rad/s 2 32
0
2 0 dt = 80 d ,
-12.7687 T = - 80 32
T = 630 s, after t = 28 s t = 658 s !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 28.
The flywheel shown has a radius of 600 mm, a mass of 144 kg, and a radius of gyration of 450 mm. An 18-kg block A is attached to a wire that is wrapped around the flywheel, and the system is released from rest. Neglecting the effect of friction, determine (a) the acceleration of block A, (b) the speed of block A after it has moved 1.8 m.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 28.
Kinematics
Kinetics
M B = ( M B )eff :
( mA g ) r
= I + ( m Aa ) r
a m A gr = mF k 2 + m Aar r
a=
mA g k mA + mF r
2
(a)
a =
(18 kg ) ( 9.81 m/s 2 )
450 mm 18 kg + (144 kg ) 600 mm
2
= 1.7836 m/s 2
or a A = 1.784 m/s 2
!
(b)
For s = 1.8 m
2 2 VA + VB + 2as
2 VA = 0 + 2 1.7836 m/s 2 (1.8 m ) = 6.42096 m 2 /s 2
(
)
VA = 2.5339 m/s
or VA = 2.53 m/s
!
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 77.
A 120-mm-diameter hole is cut as shown in a thin disk of 600-mm diameter. The disk rotates in a horizontal plane about its geometric center A at the constant rate of 480 rpm. Knowing that the disk has a mass of 30 kg after the hole has been cut, determine the horizontal component of the force exerted by the shaft on the disk at A.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 77.
Determination of mass center of disk We determine the centroid of the composite area:
xA = x1 A1 - x2 A2 or x ( A1 - A2 ) = x1 A1 - x2 A2
x =
xA1 - x2 A2 0 - (160) (60) 2 (160)(60)2 = =- = - 6.667 mm A1 - A2 (300) 2 - (60)2 (300)2 - (60) 2
Mass center G coincides with centroid C
= 480 rpm = 50.265 rad/s
a = r 2 = (6.667 10-3 m)(50.265 rad/s)2 = 16.844 m/s 2 F = ( F )eff : A = m a = (30 kg)(16.844 m/s 2 ) A = 505 N
!
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 82 .
A uniform rod of length L and mass m is supported as shown. Knowing that cable AD suddenly breaks, determine at this instant (a) the angular acceleration of the rod, (b) the tensions in cables AE and BF.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization 16, System
Chapter Solution 82.
Kinematics: a B = a A + a B/ A
aB
=
30
aA
L = 2
2L 3
L 3
L 2
aG = L 2
3 mL Fx = TAE = 2 3 mL 1 Fy = TAE + TBF - mg = - 2 2 1 1 L L mL2 M G = -TAE + TBF = 2 2 2 12
unknowns: TAE , TBF ,
Solve for,
(a)
(b) TAE =
=
3g ! 4L
mg 3mg , TBF = ! 2 8
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 84.
Two identical 4-lb slender rods AB and BC are connected by a pin at B and by the cord AC. The assembly rotates in a vertical plane under the combined effect of gravity and a 6 lb ft couple M applied to rod AB. Knowing that in the position shown the angular velocity of the assembly is zero, determine (a) the angular acceleration of the assembly, (b) the tension in cord AC.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 84.
(a)
M A = (4 lb)(0.5 ft) + (4 lb)(1.25 ft) - 6 lb ft
4 lb 1 4 lb = 2 (1 ft) 2 + (0.5 ft) 2 2 2 12 32.2 ft/s 32.2 ft/s
4 lb 4 lb + (1.25 ft)2 + [0.5 ft(0.866)]2 2 2 32.2 ft/s 32.2 ft/s
(7 - 6) lb ft = (0.26915)
= 3.7154 rad/s 2
or = 3.72 rad/s 2 (b)
!
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
M B = (4 lb)(0.25 ft) - T (1 ft)(sin 30)
= 4 lb 1 4 lb (1 ft) 2 + (1.25 ft)(0.25 ft) 2 2 12 32.2 ft/s 32.2 ft/s 4 lb (0.5 ft)(cos.30)(0.433 ft) + 2 32.2 ft/s
= 0.07246 = (0.07246)(3.7154) = 1 - T (0.5) = 0.26922 T = 1.462 lb or T = 1.462 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 85.
Two identical 4-lb slender rods AB and BC are connected by a pin at B and by the cord AC. The assembly rotates in a vertical plane under the combined effect of gravity and a couple M applied to rod AB. Knowing that in the position shown the angular velocity of the assembly is zero and the tension in cord AC is 0.45 lb, determine (a) the angular acceleration of the assembly, (b) the magnitude of the couple M.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 85.
From P16.84 ABC, BC,
M A = 7 - M = 0.26915 M B = 1 - 0.5T = 0.07246 1 - (0.5)(0.45) = 0.07246
(a)
= 10.695 rad/s 2
or = 10.70 rad/s 2
!
(b)
M = 7 - 0.2695 = 4.12 ft lb
or M = 4.12 ft lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 101.
A 10-lb uniform square plate is supported by two identical 3-lb uniform slender rods AD and BE. It is held in the position shown by rope CF. Determine, immediately after rope CF has been cut, (a) the acceleration of the plate, (b) the force exerted on the plate at point B.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 101.
(a)
ma =
10 8 32.2 12
where is angular acceleration of bar AD and bar BE about fixed axes of rotation. The plate is translating so a A = aB
Plate:
M G = 0;
R2 = - R3
2
Bar AD:
1 3 8 4 8 M D = (3 lb) ft (cos 45) - ft R2 = ft 3 32.2 12 12 12 0.7071 - 0.6667 R2 = 0.01380 (1)
Note for bar BE M E would yield 0.7071 - 0.6667 R4 = 0.01380 Therefore R4 = R2 ; R1 = R3
10 8 F = 2 R2 + (10 lb) (cos 45) = ft 32.2 12 2 R2 + 7.0711 = 0.20704 Solve eq (1) and (2) for and R2 0.6667 R2 + 0.01380 = 0.70711 2R2 - (0.20704) = - 7.0711
continued
(2)
(3) (4)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
From equations (3) and (4):
= 37.00 rad/s 2
8 8 a = ft = ft (37.00 rad/s 2 ) = 24.667 ft/s 2 10 12 or a = 24.7 ft/s 2 (b) 45 !
R2 = Now
1 ( - 7.0711 lb) + (0.20704) (37.00) = 0.2946 lb 2 FB = R2 cos 45 + (- R3 ) cos 45 = 2(0.2946) (0.707) = 0.417 lb FB = 0.417 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 104.
Gear C has a mass of 5 kg and a centroidal radius of gyration of 75 mm. The uniform bar AB has a mass of 3 kg and gear D is stationary. If the system is released from rest in the position shown, determine (a) the angular acceleration of gear C, (b) the acceleration of point B.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 104.
Kinematics:
Since gear D is fixed, we have for point E of gear C:
( a E )t = 0
But
a E = a B + a E/ B
+ (aE )t = (aB )t + (a E/B )t
0 = 0.2 AB - 0.1 C
AB =
Gear C
1 C 2
(1)
M B = (M B )eff :
Q (0.1m) = I C C
= (5 kg) (0.075 m) 2 C
Q = 0.28125 C
(2)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Bar AB and gear C
(a)
M A = (M A )eff :
WAB (0.1) + WC (0.2) - Q (0.3) = (mAB a ) 0.1 + I AB AB + (mC aB ) 0.2 - I C C
(3) g (0.1) + (5) g (0.2) - Q (0.3) = 3(0.1 AB ) 0.1 +
1 (3) (0.2)2 AB 12
+ 5(0.2 AB ) 0.2 - 5(0.075) 2 C
(1.3) g - 0.3 Q = 0.24 AB - 0.028125C Substituting for AB and Q from (2) and (1):
1 1.3 g - 0.3(0.28125 C ) = 0.24 C - 0.028125 C 2 1.3g = 0.17625 C
C = 7.3759(9.81)
C = 72.4 rad/s 2
!
C = 72.36
(b) 1 aB = 0.2 AB = 0.2 C = 0.1 C = 0.1(72.36) 2
a B = 7.24 m/s 2 !
Note: The same numerical values were obtained for c and a B in Prob. 16.103
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Problem 110.
The center of gravity G of a 3.5-lb unbalanced tracking wheel is located at a distance r = 0.9 in. from its geometric center B. The radius of the wheel is R = 3 in. and its centroidal radius of gyration is 2.2 in. At the instant shown the center B of the wheel has a velocity of 1.05 ft/s and an acceleration of 3.6 ft / s 2 , both directed to the left. Knowing that the wheel rolls without sliding and neglecting the mass of the driving yoke AB, determine the horizontal force P applied to the yoke.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 16, Solution 110.
Kinematics: Choose positive v B and a B to left
Trans. with B
+
Rotation Abt. B
=
Rolling motion
a = aB + r 2
Kinetics:
r + aB R
M C = ( M C )eff : PR - Wr = ma y r + ( max ) R + I
(
)
a r PR - mgr = m aB r + m aB + r 2 R + mk 2 B R R
(
)
2 r2 k2 v = maB +R+ + mr B R R R r
r2 + k 2 r 2 r P = mg + maB 1 + + m 2 vB 2 R R R
(1)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Substitute:
mg = 3.5 lb,
r = R= k =
m=
3.5 lb 32.2 ft/s 2
0.9 ft = 0.075 ft 12 3 ft = 0.25 ft 12 2.2 ft = 0.18333 ft 12
0.075 ft 3.5 lb P = 3.5 lb a + 2 ( B ) 0.25 ft 32.2 ft/s
( 0.075 ft )2 + ( 0.18333 ft )2 + 3.5 lb 0.75 ft v 2 1 + B 32.2 ft/s 2 ( 0.25 ft )2 ( 0.25 ft )2
2 P = 1.05 lb + 1.7693 lb/ft s 2 aB + 0.13043 lb/ft 2 s 2 vB
(
)
(
)
Substitute:
vB = 1.05 ft/s aB = 3.6 ft/s 2
; ;
vB = +1.05 ft/s aB = +3.6 ft/s 2
P = 1.05 lb + 0.17693 lb/ft s2 3.6 ft/s2 + 0.13043 lb/ft 2 s 2 (1.05 ft/s )
(
)(
) (
)
2
= 1.05 lb + 0.63695 lb + 0.14380 lb = 1.83075 lb P = 1.831 lb "
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.