PROBLEM 8.26
A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction s is the same at A and B, determine the smallest value of s
for which equilibrium is maintained.
SOLUTION
Free body: Ladder
Motion impending:
MA
FA
sNA
FB
s NB
0: W (1.25 m)
N B (6 m)
1.25W
6 2.5
NB
Fy
0: N A
Fx
0:
s NB
NA
W
NA
W
sNA
W
s N B (2.5
m)
0
(1)
s
0
s NB
1.25 sW
6 2.5 s
NB
(2)
0
Substitute for NA and NB from Eqs. (1) and (2):
sW
6
s
1.25
1.25 s2W
6 2.5 s
2.5
2
s
1.25
2
s
6
s
2
s
s
1.25
1.25 0
s
and
1.25W
6 2.5
s
0.2
5 (Discard)
s
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0.200
PROBLEM 8.30
In Problem 8.29, determine the range of values of the magnitude P of the
vertical force applied at E for which the plate will move downward.
PROBLEM 8.29 The 50-lb plate ABCD is attached at A and D to collars
that can slide on the vertical rod. Knowing that the coefficient of static
friction is 0.40 between both collars and the rod, determine whether the
plate is in equilibrium in the position shown when the magnitude of the
vertical force applied at E is (a) P = 0, (b) P = 20 lb.
SOLUTION
We shall consider the following two cases:
(1)
0 < P < 30 lb
ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0
N A = 75 lb − 2.5P
(Note: N A ≥ 0 and directed
for P ≤ 30 lb as assumed here)
ΣFx = 0: N A = N D
ΣFy = 0: FA + FD + P − 50 = 0
FA + FD = 50 − P
But:
( FA ) m = ( F0 ) m = µ s N A
= 0.40(75 − 2.5P )
= 30 − P
Plate moves if:
or
(2)
FA + FD > ( FA ) m + ( FD ) m
50 − P > (30 − P) + (30 − P)
P > 10 lb
30 lb < P < 50 lb
ΣM D = 0: − N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0
N A = 2.5P − 75
(Note: NA > and directed
for P > 30 lb as assumed)
ΣFx = 0: N A = N D
ΣFy = 0: FA + FD + P − 50 = 0
FA + FD = 50 − P
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1252
PROBLEM 8.30 (Continued)
But:
( FA ) m = ( FD ) m = µ s N A
= 0.40(2.5 P − 75)
= P − 30 lb
Plate moves if:
FA + FD > ( FA ) m + ( FD ) m
50 − P > ( P − 30) + ( P − 30)
P<
110
= 36.7 lb
3
10.00 lb < P < 36.7 lb
Thus, plate moves downward for:
(Note: For P > 50 lb, plate is in equilibrium)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1253
PROBLEM 8.36
Two 10-lb blocks A and B are connected by a slender rod of negligible weight. The
coefficient of static friction is 0.30 between all surfaces of contact, and the rod forms
an angle θ = 30°. with the vertical. (a) Show that the system is in equilibrium
when P = 0. (b) Determine the largest value of P for which equilibrium is
maintained.
SOLUTION
FBD block B:
(a)
Since P = 2.69 lb to initiate motion,
equilibrium exists with P = 0
(b)
For Pmax , motion impends at both surfaces:
ΣFy = 0: N B − 10 lb − FAB cos 30° = 0
Block B:
N B = 10 lb +
Impending motion:
3
FAB
2
(1)
FB = µs N B = 0.3N B
ΣFx = 0: FB − FAB sin 30° = 0
FAB = 2 FB = 0.6 N B
Solving Eqs. (1) and (2):
N B = 10 lb +
(2)
3
(0.6 N B ) = 20.8166 lb
2
FBD block A:
FAB = 0.6 N B = 12.4900 lb
Then
Block A:
ΣFx = 0: FAB sin 30° − N A = 0
NA =
Impending motion:
1
1
FAB = (12.4900 lb) = 6.2450 lb
2
2
FA = µs N A = 0.3(6.2450 lb) = 1.8735 lb
ΣFy = 0: FA + FAB cos 30° − P − 10 lb = 0
3
FAB − 10 lb
2
3
= 1.8735 lb +
(12.4900 lb) − 10 lb
2
= 2.69 lb
P = FA +
P = 2.69 lb
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1262