8–3. If the coefficient of static friction at A is m s= 0.4
and the collar at B is smooth so it only exerts a horizontal
force on the pipe, determine the minimum distance x so
that the bracket can support the cylinder of any mass
without slipping. Neglect the mass of the bracket.
100 mm
x
B
C
200 mm
A
8–7.
The block brake consists of a pin-connected lever and
friction block at B. The coefficient of static friction between
the wheel and the lever is ms = 0.3, and a torque of 5 N # m
is applied to the wheel. Determine if the brake can hold
the wheel stationary when the force applied to the lever is
(a) P = 30 N, (b) P = 70 N.
5N m
150 mm
50 mm
O
P
A
B
SOLUTION
200 mm
To hold lever:
a + ©MO = 0;
FB (0.15) - 5 = 0;
FB = 33.333 N
Require
NB =
33.333 N
= 111.1 N
0.3
Lever,
a + ©MA = 0;
PReqd. (0.6) - 111.1(0.2) - 33.333(0.05) = 0
PReqd. = 39.8 N
a) P = 30 N 6 39.8 N
No
Ans.
b) P = 70 N 7 39.8 N
Yes
Ans.
400 mm
.
8–56.
The uniform 6-kg slender rod rests on the top center of the
3-kg block. If the coefficients of static friction at the points
of contact are mA = 0.4, mB = 0.6, and mC = 0.3, determine
the largest couple moment M which can be applied to the
rod without causing motion of the rod.
C
800 mm
M
SOLUTION
B
Equations of Equilibrium: From FBD (a),
300 mm
+ ©F = 0;
:
x
FB - NC = 0
(1)
+ c ©Fy = 0;
NB + FC - 58.86 = 0
(2)
FC10.62 + NC10.82 - M - 58.8610.32 = 0
(3)
+ c ©Fy = 0;
NA - NB - 29.43 = 0
(4)
+ ©F = 0;
:
x
FA - FB = 0
(5)
a+ ©MO = 0;
FB 10.32 - NB 1x2 - 29.431x2 = 0
(6)
a+ ©MB = 0;
100 mm 100 mm
From FBD (b),
Friction: Assume slipping occurs at point C and the block tips, then
FC = msCNC = 0.3NC and x = 0.1 m. Substituting these values into Eqs. (1), (2), (3),
(4), (5), and (6) and solving, we have
M = 8.561 N # m = 8.56 N # m
NB = 50.83 N
NA = 80.26 N
600 mm
A
Ans.
FA = FB = NC = 26.75 N
Since 1FA2max = ms A NA = 0.4180.262 = 32.11 N 7 FA , the block does not slip.
Also, 1FB2max = ms B NB = 0.6150.832 = 30.50 N 7 FB , then slipping does not
occur at point B. Therefore, the above assumption is correct.
8–78.
The braking mechanism consists of two pinned arms and a
square-threaded screw with left and righthand threads. Thus
when turned, the screw draws the two arms together. If the
lead of the screw is 4 mm, the mean diameter 12 mm, and
the coefficient of static friction is ms = 0.35, determine the
tension in the screw when a torque of 5 N # m is applied to
tighten the screw. If the coefficient of static friction between
the brake pads A and B and the circular shaft is msœ = 0.5,
determine the maximum torque M the brake can resist.
5N·m
300 mm
200 mm M
A
B
300 mm
C
SOLUTION
l
4
b = tan-1 c
d = 6.057°,
2pr
2p162
M = 5 N # m and fs = tan-1ms = tan-1 10.352 = 19.290°. Since friction at two
screws must be overcome, then, W = 2P. Applying Eq. 8–3, we have
Frictional Forces on Screw: Here, u = tan-1 a
M = Wr tan1u + f2
5 = 2P10.0062 tan16.057° + 19.290°2
P = 879.61 N = 880 N
Ans.
Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M
is removed.
Equations of Equilibrium and Friction: Since the shaft is on the verge to rotate
about point O, then, FA = ms ¿NA = 0.5NA and FB = ms ¿NB = 0.5NB . From FBD (a),
a + ©MD = 0;
879.61 10.62 - NB 10.32 = 0
NB = 1759.22 N
From FBD (b),
a + ©MO = 0;
230.511759.222410.22 - M = 0
M = 352 N # m
Ans.
D
8–80.
Determine the horizontal force P that must be applied
perpendicular to the handle of the lever at A in order to
develop a compressive force of 12 kN on the material. Each
single square-threaded screw has a mean diameter of 25 mm
and a lead of 7.5 mm. The coefficient of static friction at all
contacting surfaces of the wedges is ms = 0.2, and the
coefficient of static friction at the screw is msœ = 0.15.
A
15⬚ C 15⬚
B
SOLUTION
Referring to the free-body diagram of wedge C shown in Fig. a, we have
+ c ©Fy = 0;
2N cos 15° - 230 .2N sin 15°4 - 12000 = 0
N = 6563 .39 N
Using the result of N and referring to the free-body diagram of wedge B shown in
Fig. b, we have
+ c ©Fy = 0;
N¿ - 6563 .39 cos 15° + 0 .2(6563 .39) sin 15° = 0
N¿ = 6000 N
+ ©Fx = 0;
:
T - 6563 .39 sin 15° - 0 .2(6563 .39) cos 15° - 0.2(6000) = 0
T = 4166 .68 N
Since the screw is being tightened, Eq. 8–3 should be used. Here,
u = tan - 1 c
L
7 .5
d = tan-1 c
d = 5.455°;
2pr
2p(12 .5)
fs = tan-1ms = tan-1(0 .15) = 8 .531°; M = P(0 .25); and W = T = 4166.68N. Since
M must overcome the friction of two screws,
M = 23Wr tan (fs + u)4
P(0 .25) = 234166 .68(0 .0125) tan (8 .531° + 5.455°)4
P = 104 N
Ans.
250 mm
8–91.
A cable is attached to the plate B of mass MB, passes over a fixed peg at C, and is attached to
the block at A. Using the coefficients of static friction shown, determine the smallest mass of
block A so that it will prevent sliding motion of B down the plane.
Given:
MB
20 kg
PA
0.2
T
30 deg
PB
0.3
g
9.81
m
PC
0.3
2
s
Solution:
Iniitial guesses:
T1
1N
T2
1 N NA
1N
Given
Block A:
6F x = 0;
T1 P A NA MA g sin T 6F y = 0;
NA MA g cos T 0
0
Plate B:
6F x = 0;
6F y = 0;
Peg C:
§ T1 ·
¨ ¸
¨ T2 ¸
¨ NA ¸
¨ ¸
¨ NB ¸
¨M ¸
© A¹
MA
T2 MB g sin T P B NB P A NA
NB NA MB g cos T T2
P CS
T1 e
Find T1 T2 NA NB MA
2.22 kg
Ans.
0
0
NB
1N
MA
1 kg
8–102.
The 20-kg motor has a center of gravity at G and is pinconnected at C to maintain a tension in the drive belt.
Determine the smallest counterclockwise twist or torque M
that must be supplied by the motor to turn the disk B if
wheel A locks and causes the belt to slip over the disk. No
slipping occurs at A. The coefficient of static friction
between the belt and the disk is ms = 0.35.
M
A
B
50 mm
G
50 mm
C
100 mm
SOLUTION
Equations of Equilibrium: From FBD (a),
a + ©MC = 0;
T2 11002 + T1 12002 - 196.211002 = 0
(1)
M + T1 10.052 - T2 10.052 = 0
(2)
From FBD (b),
a + ©MO = 0;
Frictional Force on Flat Belt: Here, b = 180° = p rad. Applying Eq. 8–6,
T2 = T1 emb, we have
T2 = T1 e0.3p = 3.003T1
(3)
Solving Eqs. (1), (2), and (3) yields
M = 3.93 N # m
T1 = 39.22 N
T2 = 117.8 N
Ans.
150 mm
8–103.
Blocks A and B have a mass of 100 kg and 150 kg,
respectively. If the coefficient of static friction between A
and B and between B and C is ms = 0.25 and between the
ropes and the pegs D and E m¿ s = 0.5 determine the
smallest force F needed to cause motion of block B if
P = 30 N.
E
D
A
C
SOLUTION
P
Assume no slipping between A and B.
Peg D :
T2 = T1 emb;
p
FAD = 30 e0.5( 2 ) = 65.80 N
Block B :
+ ©Fx = 0;
:
- 65.80 - 0.25 NBC + FBE cos 45° = 0
+ c ©Fy = 0;
NBC - 981 + FBE sin 45° - 150 (9.81) = 0
FBE = 768.1 N
NBC = 1909.4 N
Peg E :
T2 = T1emb;
3p
F = 768.1e0.5( 4 ) = 2.49 kN
Note: Since B moves to the right,
(FAB)max = 0.25 (981) = 245.25 N
p
245.25 = Pmax e0.5( 2 )
Pmax = 112 N 7 30 N
Hence, no slipping occurs between A and B as originally assumed.
45⬚
B
Ans.
F
4000 N. If µs = 0.35,
18.75 mm
 3
3
2
 50  25 
= 3 (0.35)(4000)  2
2
 50  25 
50 mm
P = 4000 N
25 mm
= 54 444 N  mm
M = 54.4 N  m
Ans.
8–116.
A 200-mm diameter post is driven 3 m into sand for which
ms = 0.3. If the normal pressure acting completely around the
post varies linearly with depth as shown, determine the
frictional torque M that must be overcome to rotate the post.
M
200 mm
3m
SOLUTION
Equations of Equilibrium and Friction: The resultant normal force on the post is
1
N = 1600 + 021321p210.22 = 180p N. Since the post is on the verge of rotating,
2
F = ms N = 0.31180p2 = 54.0p N.
a + ©MO = 0;
M - 54.0p10.12 = 0
M = 17.0 N # m
Ans.
600 Pa
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 horizontal 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.
W
P
A
r
SOLUTION
B
+ ©F = 0;
:
x
(RA)x - P = 0
(RA)x = P
+ c ©Fy = 0;
(RA)y - W = 0
(RA)y = W
a + ©MB = 0;
P(r cos fA + r cos fB) - W(aA + aB) = 0
(1)
Since fA and fB are very small, cos fA - cos fB = 1. Hence, from Eq. (1)
P =
W(aA + aB)
2r
(QED)