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
*8–8.
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. = 34.26 N
a) P = 30 N 6 34.26 N
No
Ans.
b) P = 70 N 7 34.26 N
Yes
Ans.
400 mm
8–23.
A 35-kg disk rests on an inclined surface for which ms = 0.2.
Determine the maximum vertical force P that may be
applied to link AB without causing the disk to slip at C.
P
200 mm
200 mm
C
SOLUTION
30°
Equations of Equilibrium: From FBD (a),
a + ©MB = 0;
P16002 - A y 19002 = 0
A y = 0.6667P
From FBD (b),
+ c ©Fy = 0
NC sin 60° - FC sin 30° - 0.6667P - 343.35 = 0
(1)
a + ©MO = 0;
FC12002 - 0.6667P12002 = 0
(2)
Friction: If the disk is on the verge of moving, slipping would have to occur at
point C. Hence, FC = ms NC = 0.2NC . Substituting this value into Eqs. (1) and (2)
and solving, we have
P = 182 N
NC = 606.60 N
Ans.
A
300 mm
600 mm
B
8–46.
The beam AB has a negligible mass and thickness and is
subjected to a triangular distributed loading. It is supported
at one end by a pin and at the other end by a post having a
mass of 50 kg and negligible thickness. Determine the
minimum force P needed to move the post. The coefficients
of static friction at B and C are mB = 0.4 and mC = 0.2,
respectively.
800 N/m
A
B
2m
400 mm
4
300 mm
C
SOLUTION
Member AB:
a + ©MA = 0;
4
- 800a b + NB (2) = 0
3
NB = 533.3 N
Post:
Assume slipping occurs at C; FC = 0.2 NC
a + ©MC = 0;
4
- P(0.3) + FB(0.7) = 0
5
+ ©F = 0;
:
x
4
P - FB - 0.2NC = 0
5
+ c ©Fy = 0;
3
P + NC - 533.3 - 50(9.81) = 0
5
P = 355 N
Ans.
NC = 811.0 N
FB = 121.6 N
(FB)max = 0.4(533.3) = 213.3 N 7 121.6 N
(O.K.!)
P
5
3
8–47.
The beam AB has a negligible mass and thickness and is
subjected to a triangular distributed loading. It is supported
at one end by a pin and at the other end by a post having a
mass of 50 kg and negligible thickness. Determine the two
coefficients of static friction at B and at C so that when the
magnitude of the applied force is increased to P = 150 N,
the post slips at both B and C simultaneously.
800 N/m
A
B
2m
400 mm
4
300 mm
C
SOLUTION
Member AB:
a + ©MA = 0;
4
- 800a b + NB (2) = 0
3
NB = 533.3 N
Post:
+ c ©Fy = 0;
3
NC - 533.3 + 150a b - 50(9.81) = 0
5
NC = 933.83 N
a + ©MC = 0;
4
- (150)(0.3) + FB (0.7) = 0
5
FB = 51.429 N
+ ©F = 0;
:
x
4
(150) - FC - 51.429 = 0
5
FC = 68.571 N
mC =
FC
68.571
= 0.0734
=
NC
933.83
Ans.
mB =
FB
51.429
= 0.0964
=
NB
533.3
Ans.
P
5
3
8–65.
The coefficient of static friction between wedges B and C is
ms = 0.6 and between the surfaces of contact B and A and
C and D, ms ¿ = 0.4. If the spring is compressed 200 mm
when in the position shown, determine the smallest force P
needed to move wedge C to the left. Neglect the weight of
the wedges.
k
A
15
B
15
C
15
D
SOLUTION
Wedge B:
+ ©F = 0;
:
x
NAB - 0.6NBC cos 15° - NBC sin 15° = 0
+ c ©Fy = 0;
NBC cos 15° - 0.6NBC sin 15° - 0.4NAB - 100 = 0
NBC = 210.4 N
NAB = 176.4 N
Wedge C:
+ c ©Fy = 0;
NCD cos 15° - 0.4NCD sin 15° + 0.6(210.4) sin 15° - 210.4 cos 15° = 0
NCD = 197.8 N
+ ©F = 0;
:
x
197.8 sin 15° + 0.4(197.8) cos 15° + 210.4 sin 15° + 0.6(210.4) cos 15° - P = 0
P = 304 N
Ans.
500 N/ m
P
8–66.
The coefficient of static friction between the wedges B and
C is ms = 0.6 and between the surfaces of contact B and A
and C and D, ms ¿ = 0.4. If P = 50 N, determine the largest
allowable compressionof the spring without causing wedgeC
to move to the left. Neglect the weight of the wedges.
k
A
15
B
15
C
15
D
SOLUTION
Wedge C:
+ ©F = 0;
:
x
(NCD + NBC) sin 15° + (0.4NCD + 0.6NBC) cos 15° - 50 = 0
c + ©Fy = 0;
(NCD - NBC) cos 15° + ( -0.4NCD + 0.6NBC) sin 15° = 0
NBC = 34.61 N
NCD = 32.53 N
Wedge B:
+ ©F = 0;
:
x
NAB - 0.6(34.61) cos 15° - 34.61 sin 15° = 0
NAB = 29.01 N
c + ©Fy = 0;
34.61 cos 15° - 0.6(34.61) sin 15° - 0.4(29.01) - 500x = 0
x = 0.03290 m = 32.9 mm
Ans.
500 N/m
P
*8–92.
The simple band brake is constructed so that the ends of the
friction strap are connected to the pin at A and the lever arm
at B. If the wheel is subjected to a torque of M = 80 lb # ft,
and the minimum force P = 20 lb is needed to apply to the
lever to hold the wheel stationary, determine the coefficient
of static friction between the wheel and the band.
M ⫽ 80 lb ⭈ ft
O
20⬚
45⬚
1.25 ft
B
A
1.5 ft
SOLUTION
P
Equations of Equilibrium: Write the moment equation of equilibrium about
point A by referring to the FBD of the lever shown in Fig. a,
a + ©MA = 0;
TB sin 45°(1 .5) - 20(4 .5) = 0
TB = 84.85 lb
Using this result to write the moment equation of equilibrium about point 0 by
referring to the FBD of the wheel shown in Fig. b,
a + ©MO = 0;
3 ft
TA(1 .25) + 80 - 84 .85(1 .25) = 0
Frictional Force on Flat Belt: Here, b = a
TA = 20.85 lb
245°
49
bp =
p, T1 = TA = 20.85 lb and
180°
36
T2 = TB = 84.85 lb. Applying Eq. 8–6,
T2 = T1emb
49
84 .85 = 20 .85em(36)p
49
em(36)p = 4 .069
49
In em(36)p = In 4 .069
ma
49
bp = In 4 .069
36
m = 0 .328
Ans.
8–93.
The simple band brake is constructed so that the ends of the
friction strap are connected to the pin at A and the lever
arm at B. If the wheel is subjected to a torque of
M = 80 lb # ft, determine the smallest force P applied to the
lever that is required to hold the wheel stationary. The
coefficient of static friction between the strap and wheel is
ms = 0.5.
M
O
20
1.5 ft
T1(1.25) + 80 - T2(1.25) = 0
p
T2 = T1e0.5(245°)(180°) = 8.4827T1
T1 = 8.553 lb
T2 = 72.553 lb
P = 17.1 lb
3 ft
P
Solving;
a + ©MA = 0;
B
A
b = 20° + 180° + 45° = 245°
T2 = T1emb;
45
1.25 ft
SOLUTION
a + ©MO = 0;
80 lb ft
- 72.553(sin 45°)(1.5) - 4.5P = 0
Ans.
8–130.
The hand cart has wheels with a diameter of 80 mm. If a
crate having a mass of 500 kg is placed on the cart so that
each wheel carries an equal load, determine the horizontal
force P that must be applied to the handle to overcome the
rolling resistance. The coefficient of rolling resistance is
2 mm. Neglect the mass of the cart.
P
SOLUTION
P L
Wa
r
= 50019.812a
P = 245 N
2
b
40
Ans.
8–141.
The jacking mechanism consists of a link that has a squarethreaded screw with a mean diameter of 0.5 in. and a lead of
0.20 in., and the coefficient of static friction is ms = 0.4.
Determine the torque M that should be applied to the
screw to start lifting the 6000-lb load acting at the end of
member ABC.
6000 lb
C
7.5 in.
B
M
10 in.
D
SOLUTION
A
-1
a = tan
10
a b = 21.80°
25
a + ©MA = 0;
20 in.
- 6000 (35) + FBD cos 21.80° (10) + FBD sin 21.80° (20) = 0
FBD = 12 565 lb
fs = tan-1 (0.4) = 21.80°
u = tan-1 a
0.2
b = 7.256°
2p (0.25)
M = Wr tan (u + f)
M = 12 565 (0.25) tan (7.256° + 21.80°)
M = 1745 lb # in = 145 lb # ft
Ans.
15 in.
10 in.