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 m s
=
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.
SOLUTION
To hold lever: a + © M
O
=
0;
F
B
(0.15)
-
5
=
0;
Require
F
B
=
33.333 N
N
B
=
33.333 N
0.3
= 111.1 N
Lever,
P
Reqd.
(0.6) 111.1(0.2) 33.333(0.05) = 0 a + © M
A
= 0;
P
Reqd.
b)
= 39.8 N a) P = 30 N 6 39.8 N No
P = 70 N 7 39.8 N Yes
5 N m
50 mm
A
150 mm
O
B
200 mm 400 mm
P
A n s .
A n s .

*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 m s
=
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) .
SOLUTION
To hold lever: a + © M
O
= 0; F
B
(0.15) + 5 = 0;
Require
N
B
=
33.333 N
0.3
= 111.1 N
F
B
= 33.333 N
Lever,
P
Reqd.
(0.6) 111.1(0.2) + 33.333(0.05) = 0 a + © M
A
= 0;
P
Reqd.
= 34.26 N a) P = 30 N 6 34.26 N No b) P = 70 N 7 34.26 N Yes
5 N m
50 mm
A
150 mm
O
B
200 mm 400 mm
P
A n s .
A n s .

8–23.
A 35-kg disk rests on an inclined surface for which m s
=
0.2.
Determine the maximum vertical force P that may be applied to link AB without causing the disk to slip at C .
200 mm
300 mm
P
600 mm
SOLUTION
Equations of Equilibrium: From FBD (a), a + © M
B
=
0;
P 1
600
2 A y
1
900
2 =
0
From FBD (b),
+ c © F y
= 0 a + © M
O
= 0;
A y
=
0.6667
P
N
C sin 60° F
C sin 30° 0.6667
P 343.35
= 0
F
C
1 200 2 0.6667
P 1 200 2 = 0
(1)
(2)
Friction: If the disk is on the verge of moving, slipping would have to occur at point C . Hence, F
C
= m s
N
C
= 0.2
N
C
.
Substituting this value into Eqs. (1) and (2) and solving, we have
P =
182 N
N
C
=
606.60 N
A n s .
C
200 mm
A
30
°
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 m
B
= 0.4
= 0.2, respectively.
SOLUTION
Member AB: a + © M
A
= 0; 800 a
4
3 b + N
B
(2) = 0
N
B
= 533.3 N
Post:
Assume slipping occurs at C ; F
C
= 0.2
N
C a + © M
C
= 0; -
4
5
P (0.3) + F
B
(0.7) = 0
F x
= 0;
4
5
P F
B
0.2
N
C
= 0
+ c © F y
= 0;
3
5
P + N
C
533.3
50(9.81) = 0
P = 355 N
N
C
= 811.0 N
F
B
= 121.6 N
( F
B
) max
= 0.4(533.3) = 213.3 N 7 121.6 N
A
A n s .
(O.K.
!
)
2 m
400 mm
800 N
/ m
B
5
4
3
P
300 mm
C

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.
SOLUTION
Member AB: a + © M
A
= 0; 800 a
4
3 b + N
B
(2) = 0
N
B
= 533.3 N
Post:
+ c © F y
= 0; N
C
533.3
+ 150 a
3
5 b - 50(9.81) = 0
N
C
= 933.83 N a + © M
C
= 0; -
4
5
(150)(0.3) + F
B
(0.7) = 0
F
B
= 51.429 N
+ ©
F x
= 0;
4
5
(150) F
C
51.429
= 0
F
C
=
68.571 N m
C
=
F
C
N
C
=
68.571
933.83
= 0.0734
m
B
=
F
B
N
B
=
51.429
533.3
= 0.0964
A
A n s .
A n s .
2 m
400 mm
800 N
/ m
B
5
4
3
P
300 mm
C

8–65.
The coefficient of static friction between wedges B and C is m
C s
= 0.6
and D , and between the surfaces of contact m s
¿ =
0.4.
B and A and
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.
SOLUTION
Wedge B :
: F x
=
0;
+ c © F y
=
0;
N
AB
-
0.6
N
BC cos 15°
N
BC
sin 15°
=
0
N
BC cos 15°
-
0.6
N
BC sin 15°
-
0.4
N
AB
-
100
=
0
N
BC
=
210.4 N
N
AB
=
176.4 N
Wedge C :
+ c © F y
= 0;
+ ©
F x
= 0;
N
CD cos 15° 0.4
N
CD sin 15° + 0.6(210.4) sin 15° 210.4 cos 15° = 0
N
CD
= 197.8 N
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 A n s .
k 500 N
/ m
A
15
B
D
15
15
C P

8–66.
The coefficient of static friction between the wedges B and
C is and m
C s
= and
0.6
D , and between the surfaces of contact m s
¿ =
0.4.
If
P =
50 N,
B and A determine the largest allowable compressionof the spring without causing wedge C to move to the left. Neglect the weight of the wedges.
SOLUTION
Wedge C :
+ ©
F x
= 0; c + © F y
= 0;
( N
CD
+ N
BC
) sin 15° + (0.4
N
CD
+ 0.6
N
BC
) cos 15° 50 = 0
( N
CD
N
BC
) cos 15° + ( 0.4
N
CD
+ 0.6
N
BC
) sin 15° = 0
N
BC
= 34.61 N
N
CD
= 32.53 N
Wedge B :
: F x
=
0; c + © F y
=
0;
N
AB
-
0.6(34.61) cos 15°
-
34.61 sin 15°
=
0
N
AB
=
29.01 N
34.61 cos 15°
-
0.6(34.61) sin 15°
-
0.4(29.01)
-
500 x =
0 x =
0.03290 m
=
32.9 mm A n s .
k 500 N / m
A
15
B
D
15
15
C P

*8–92.
The s imple band brake i s con s tructed s o that the end s of the friction s trap are connected to the pin at A and the le v er arm
# at B . If the wheel i s s ubjected to a torque of M
=
80 lb ft, and the minimum force P = 20 lb i s needed to apply to the le v er to hold the wheel s tationary, determine the coefficient of s tatic friction between the wheel and the band.
SOLUTION
Eq u at i ons of Eq ui l i br iu m : Write the moment equation of equilibrium about point A by referrin g to the FBD of the le v er s hown in Fi g .
a , a
+ ©
M
A =
0 ; T
B s in 45°(1 .5)
-
20(4 .5)
=
0 T
B =
84 .85 lb
U s in g thi s re s ult to write the moment equation of equilibrium about point 0 by referrin g to the FBD of the wheel s hown in Fi g .
b , a
+ ©
M
O =
0 ; T
A
(1 .25)
+
80
-
84 .85(1 .25)
=
0 T
A =
20 .85 lb
Fr i ct i onal Force on Flat Belt : Here, b
= a
245°
180° b p
=
T
2
= T
B
= 84.85 lb. Applyin g Eq. 8–6,
T
2
=
T
1 e mb
84 .85
=
20 .85
e m (
49
36
) p e m (
49
36
) p
=
4 .069
In e m (
49
36
) p
=
In 4 .069
49
36 p , T
1 =
T
A =
20 .85 lb and m a
49
36 b p
=
In 4 .069
m
=
0 .328
Ans.
20 ⬚
M
⫽
80 lb
⭈
ft
O
45
⬚
1.25 ft
A
B
1.5 ft 3 ft
P

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
M = 80 lb
# . If the wheel is subjected to a torque of 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 m s
=
0.5.
SOLUTION b =
20°
+
180°
+
45°
=
245° a + © M
O
=
0;
T
2
= T
1 e mb
;
T
1
(1.25)
+
80
T
2
(1.25)
=
0
T
2
= T
1 e 0.5(245°)( p
180°
) =
8.4827
T
1
Solving;
T
1
= 8.553 lb
T
2
= 72.553 lb a + © M
A
= 0;
P = 17.1 lb
72.553(sin 45°)(1.5) 4.5
P = 0
A n s .
20
M 80 lb ft
O
45
1.25 ft
A
B
1.5 ft 3 ft
P

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.
SOLUTION
P L
Wa r
= 500 1 9.81
2a
2
40 b
P =
245 N A n s .
P

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 m s
=
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 .
SOLUTION
A a = tan
1 a
10
25 b = 21.80° a + © M
A
= 0; 6000 (35) + F
BD
cos 21.80° (10) + F
BD
sin 21.80° (20) = 0
F
BD
= 12 565 lb f s
= tan
1 (0.4) = 21.80° u = tan
1 a
0.2
2 p (0.25) b = 7.256°
M = Wr tan ( u + f )
M = 12 565 (0.25) tan (7.256° + 21.80°)
M = 1745 lb
# in = 145 lb
# ft A n s .
20 in.
B
15 in.
M
6000 lb
C
10 in.
7.5 in.
10 in.
D