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9.14 Problem The box is stationary on the inclined surface. The coefcient of static friction between the box and the surface is s . (a) (b) If the mass of the box is 10 kg, D 20 , D 30 , and s D 0.24, what force T is necessary to start the box sliding up the surface? Show that the force T necessary to start the box sliding up the surface is a minimum when tan D s .
T
Solution:
T mg
y
f N
x
D 20
s
D 0.24
m D 10 kg g D 9.81 m/s2 Fx : Fy : (a) T cos C f C mg sin D 0 N C T sin
s
mg cos D 0 N
D 30 , f D
Substituting the known values and solving, we get T D 56.5 N, N D 64.0 N, f D 15.3 N Solving the 2nd equilibrium eqn for N and substituting for f f D s N in the rst eqn, we get T cos C
s mg cos s T sin
C mg sin D 0
Differentiating with respect to , we get dT T sin D d cos C Setting tan D cos sin s
s
dT D 0, we get d
s
c 2008 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.
693
Problem 9.27 The ladder and the person weigh 30 lb and 180 lb, respectively. The center of mass of the 12-ft ladder is at its midpoint. The angle D 30 . Assume that the wall exerts a negligible friction force on the ladder. (a) If x D 4 ft, what is the magnitude of the friction force exerted on the ladder by the oor? (b) What minimum coefcient of static friction between the ladder and the oor is necessary for the person to be able to climb to the top of the ladder without slipping?
a
x
Solution:
(a) Assume no slipping occurs D 30 , x D 4 ft Fx : fB Fy : NB NA D 0 210 lb D 0 30 lb 6 ft sin 180 lb x D 0
NA
MB : NA 12 ft cos Solving (b)
180 lb
fB D 77.9 lb, NB D 210 lb
At the top of the ladder D 30 , x D 6 ft Fx : fB Fy : NB NA D 0
30 lb fB x
NB
210 lb D 0 30 lb 6 ft sin 180 lb x D 0
MB : NA 12 ft cos fB D
s NB
)
s
D 0.536
c 2008 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.
701
Problem 9.37 The mass of block B is 8 kg. The coefcient of static friction between the surfaces of the clamp and the block is s D 0.2. When the clamp is aligned as shown, what minimum force must the spring exert to prevent the block from slipping out?
200 mm
45 160 mm
B
100 mm
Solution: The free-body diagram of the block when slip is impending is shown. From the equilibrium equation
s FT
FT
sFT
C
s
FT C W cos
W cos D 0,
W
s (FT + Wcos )
we obtain FT D w1 2
s s
FT +Wcos 100 mm
cos
D
8 9.81 1 0.2 2 0.2
cos 45
sFT
FS FT
160 200 mm mm
D 111 N. The free-body diagram of the upper arm of the clamp is shown. Summing moments about the upper end, 0.16Fs C 0.1
s FT
0.36FT D 0,
the force exerted by the spring is Fs D 0.36FT 0.1 0.16
s FT
D
[0.36
0.1 0.2 ]111 0.16
D 236 N.
Problem 9.38 By altering its dimensions, redesign the clamp in Problem 9.37 so that the minimum force the spring must exert to prevent the block from slipping out is 180 N. Draw a sketch of your new design. Solution: This problem does not have a unique solution.
c 2008 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.
707
Problem 9.51 The table weighs 50 lb and the coefcient of static friction between its legs and the inclined surface is 0.7. (a) If you apply a force at A parallel to the inclined surface to push the table up the inclined surface, will the table tip over before it slips? If not, what force is required to start the table moving up the surface? (b) If you apply a force at B parallel to the inclined surface to push the table down the inclined surface, will the table tip over before it slips? If not, what force is required to the start table moving down the surface? Solution:
(a) Assume the table does not tip F% : F f1 f2 50 lb sin 20 D0
B 32 in A 28 in
23 in 23 in 20
20
F
F- : N1 C N2 M2 : F 32 in 50 lb cos 20 D 0
f2 50 lb
N1 46 in
C 50 lb cos 20 23 in C 50 lb sin 20 28 in D 0 f1 D 0.7N1 , f2 D 0.7N2 Solving we nd F D 50 lb, N1 D 0.874 lb, N2 D 47.9 lb
f1
N2
N1 F
Since N1 < 0 we conclude that the table tips before it slips (b) Assume the table does not tip F% : F C f1 C f2 50 lb sin 20 D 0
20
F- : N1 C N2 M2 : F 32 in
50 lb cos 20 D 0 N1 46 in
f2 50 lb
C 50 lb cos 20 23 in C 50 lb sin 20 28 in D 0 f1 D 0.7N1 , f2 D 0.7N2 Solving we nd F D 15.79 lb, N1 D 44.8 lb, N2 D 2.10 lb Since N2 > 0 we conclude that the table does not tip, and F D 15.79 lb
f1
N2
N1
c 2008 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.
717
Problem 9.59 The frame is supported by the normal and friction forces exerted on the plates at A and G by the xed surfaces. The coefcient of static friction at A is s D 0.6. Will the frame slip at A when it is subjected to the loads shown?
1m
1m
1m 6 kN
A
B
C 1m
8 kN 1m
G
E
D
Solution: The strategy is to write the equilibrium equations and solve for the unknown. The complete structure as a free body: Denote the normal forces as A and G, and the friction forces as fA and fG . The sum of forces:
(1) (2) Fx D Fy D fA C fG C 8 D 0, ACG 6 D 0.
Cy D 18 kN, Cx D 13 kN, B D 36 kN, Dx D 5 kN, Dy D 18 kN, E D 36 kN,
The elements as free bodies: Element ABC: (See Figure) (3) (4) (5) MB D C1A C 2Cy Fx D Fy D 12 D 0.
GD fA C Cx D 0,
18 kN,
fG D 5 kN. A B C Cy 6 D 0.
Element BE: (6) Fy D B Element CD: (7) (8) (9) MC D C2Dx Fx D Dy C 8 D 0, E D 0.
fA A B E fG G
B
6 kN C X CY CX CY
E
DX DY
DY DX
8 kN
Cx C Dx C 8 D 0 Cy D 0.
Fy D Dy Element DEG:
The assumed directions are shown in the Figure; a negative sign means that the result is opposite to the assumed direction. The magnitude of the coefcient of static friction for the reaction at A required to hold the frame in equilibrium is D fA D 0.5417. A
s
(10) (11) (12)
MD D
2G
ED0 Dx D 0
sA
Fx D fG Fy D G
Dy C E D 0.
Since this is less than the known value, slip at A.
D 0.6, the frame will not
These twelve equations in ten unknowns can be solved by iteration or by back substitution. The results in detail: AD 24 kN,
fA D 13 kN,
c 2008 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.
723
Problem 9.60 The frame is supported by the normal and friction forces exerted on the plate at A by the wall. (a) What is the magnitude of the friction force exerted on the plate at A? (b) What is the minimum coefcient of static friction at A necessary for the structure to remain in equilibrium?
1m
E 6 kN D C
1m
A B 2m 2m 1m
Solution: Draw a free body diagram of each member and write
the corresponding equilibrium equations Fx : N C Bx D 0 Fy : f C By C Cy D 0 MA : 2By C 4Cy D 0 Fx : Ex C Dx Fy : Ey C Dy ME : 1Dy C 1Dx Fx : Fy : MD : Unknowns: N, f, Bx , By , Cy , Dx , Dy , Ex , Ey . We have 9 eqns in 9 unknowns. Solving, we get (a) fD 8 kN (friction acts down) Dx D 0 Dy 3Cy Cy 6D0 Bx D 0 By D 0 2By 2Bx D 0 (1)
y BY f A B CY C x 2m
(2)
N
(3)
BX 2m
(4) (5) (6) (7) (8) (9)
y
EY
1m
1m EX DY D DX 1m B x BX BY 1m
E
4 6 D0
Also N D 15 kN Bx D Cy D Dx D 0, Ex D (b)
MIN
y 3m DX DY CY 1m
6 kN x
15 kN, By D 16 kN 8 kN (opposite the direction assumed) Dy D 2 kN
15 kN, Ey D 14 kN D jf j D 0.533 N
724
c 2008 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.

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