8 8–61. If the spring is compressed 60 mm and the coefficient of static friction between the tapered stub S and the slider A is mSA = 0.5, determine the horizontal force P needed to move the slider forward. The stub is free to move without friction within the fixed collar C . The coefficient of static friction between A and surface B is mAB = 0.4. Neglect the weights of the slider and stub.
k 300 N/m C
A S
B 30 P
8–58. Determine the largest angle u that will cause the wedge to be self-locking regardless of the magnitude of horizontal force P applied to the blocks. The coefficient of static friction between the wedge and the blocks is ms = 0.3 . Neglect the weight of the wedge.
P P
u
Prob. 8–58
8–59. If the beam AD is loaded as shown, determine the horizontal force P which must be applied to the wedge in order to remove it from under the beam. The coefficients of static friction at the wedge’s top and bottom surfaces are mCA = 0.25 and mCB = 0.35, respectively. If P = 0, is the wedge self-locking? Neglect the weight and size of the wedge and the thickness of the beam.
3 m
A P
10 4 kN/m
C B 4 m
D
Prob. 8–59
*8–60. The wedge has a negligible weight and a coefficient of static friction ms = 0.35 with all contacting surfaces.
Determine the largest angle u so that it is “self-locking.”
This requires no slipping for any magnitude of the force P applied to the joint.
u u
P P
––2 ––
2
Prob. 8–60
8
*8–68. If the clamping force on the boards is 600 lb, determine the required magnitude of the couple forces that must be applied perpendicular to the lever AB of the clamp at A and B in order to loosen the screw. The single square-threaded screw has a mean diameter of 1 in. and a lead of 0.25 in. The coefficient of static friction is ms = 0.3 .
6 in.
6 in.
A B
Probs. 8–67/68 8–69. The column is used to support the upper floor. If a force F= 80 N is applied perpendicular to the handle to tighten the screw, determine the compressive force in the column. The square-threaded screw on the jack has a coefficient of static friction of ms = 0.4, mean diameter of 25 mm, and a lead of 3 mm.
8–70. If the force F is removed from the handle of the jack in Prob. 8–69 , determine if the screw is self-locking.
0.5 m F
Probs. 8–69/70
8–71. If the clamping force at G is 900 N, determine the horizontal force F that must be applied perpendicular to the handle of the lever at E . The mean diameter and lead of both single square-threaded screws at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3 .
*8–64. Determine the largest weight of the wedge that can be placed between the 8-lb cylinder and the wall without upsetting equilibrium. The coefficient of static friction at A and C is ms= 0.5 and at B , m=s = 0.6.
30
0.5 ft
B
A
C
Prob. 8–64
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 , m=s= 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.
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 , m=s = 0.4. If P = 50 N, determine the smallest allowable compression of the spring without causing wedge C to move to the left. Neglect the weight of the wedges.
k 500 N/m
P A B
15C 15 15
D
Probs. 8–65/66
8–67. If couple forces of F= 10 lb are applied perpendicular to the lever of the clamp at A and B , determine the clamping force on the boards. The single square-threaded screw of the clamp has a mean diameter of 1 in. and a lead of 0.25 in.
The coefficient of static friction is ms = 0.3 .
8 8–75. The shaft has a square-threaded screw with a lead of 8 mm and a mean radius of 15 mm. If it is in contact with a plate gear having a mean radius of 30 mm, determine the resisting torque M on the plate gear which can be overcome if a torque of 7 N
#
m is applied to the shaft. The coefficient of static friction at the screw is mB = 0.2. Neglect friction of the bearings located at A and B .15 mm
M
30 mm B
A
7 Nm
Prob. 8–75
*8–76. The square-threaded screw has a mean diameter of 20 mm and a lead of 4 mm. If the weight of the plate A is 5 lb, determine the smallest coefficient of static friction between the screw and the plate so that the plate does not travel down the screw when the plate is suspended as shown.
A
Prob. 8–76 *8–72. If a horizontal force of F = 50 N is applied
perpendicular to the handle of the lever at E, determine the clamping force developed at G . The mean diameter and lead of the single square-threaded screw at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.
200 mm
E D G C
200 mm
125 mm B
A
Probs. 8–71/72
8–73. A turnbuckle, similar to that shown in Fig. 8–17 , is used to tension member AB of the truss. The coefficient of the static friction between the square threaded screws and the turnbuckle is ms = 0.5. The screws have a mean radius of 6 mm and a lead of 3 mm. If a torque of M = 10 N
#
m is applied to the turnbuckle, to draw the screws closer together, determine the force in each member of the truss.No external forces act on the truss.
8–74. A turnbuckle, similar to that shown in Fig. 8–17 , is used to tension member AB of the truss. The coefficient of the static friction between the square-threaded screws and the turnbuckle is ms = 0.5. The screws have a mean radius of 6 mm and a lead of 3 mm. Determine the torque M which must be applied to the turnbuckle to draw the screws closer together, so that the compressive force of 500 N is developed in member BC .
M
4 m
C
3 m
A
B D
Prob. 8–74
8
8–79. If a horizontal force of P = 100 N is applied perpendicular to the handle of the lever at A , determine the compressive force F exerted 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 ms= = 0.15.
*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 m=s = 0.15.
A
B
250 mm 15 C 15
Probs. 8–79/80
8–81. Determine the clamping force on the board A if the screw of the “C” clamp is tightened with a twist of M = 8 N
#
m. The single square-threaded screw has a mean radius of 10 mm, a lead of 3 mm, and the coefficient of static friction is ms = 0.35.8–82. If the required clamping force at the board A is to be 50 N, determine the torque M that must be applied to the handle of the “C” clamp to tighten it down. The single square-threaded screw has a mean radius of 10 mm, a lead of 3 mm, and the coefficient of static friction is ms= 0.35.
M
A
Probs. 8–81/82 8–77. The fixture clamp consist of a square-threaded
screw having a coefficient of static friction of ms = 0.3, mean diameter of 3 mm, and a lead of 1 mm. The five points indicated are pin connections. Determine the clamping force at the smooth blocks D and E when a torque of M = 0.08 N
#
m is applied to the handle of the screw.40 mm E 45 B
30 mm
30 mm
40 mm 40 mm
D
C A
M = 0.08 Nm Prob. 8–77
8–78. The braking mechanism consists of two pinned arms and a square-threaded screw with left- and right-hand 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 m=s = 0.5, determine the maximum torque M the brake can resist.300 mm
300 mm 5Nm
200 mm
C
A B
M
D
Prob. 8–78
8
8.5 Frictional Forces on Flat Belts
Whenever belt drives or band brakes are designed, it is necessary to determine the frictional forces developed between the belt and its contacting surface. In this section we will analyze the frictional forces acting on a flat belt, although the analysis of other types of belts, such as the V-belt, is based on similar principles.
Consider the flat belt shown in Fig. 8–18 a , which passes over a fixed curved surface. The total angle of belt to surface contact in radians is b , and the coefficient of friction between the two surfaces is m. We wish to determine the tension T
2in the belt, which is needed to pull the belt counterclockwise over the surface, and thereby overcome both the frictional forces at the surface of contact and the tension T
1in the other end of the belt. Obviously, T
27 T
1.
Frictional Analysis. A free-body diagram of the belt segment in contact with the surface is shown in Fig. 8–18 b . As shown, the normal and frictional forces, acting at different points along the belt, will vary both in magnitude and direction. Due to this unknown distribution, the analysis of the problem will first require a study of the forces acting on a differential element of the belt.
A free-body diagram of an element having a length ds is shown in Fig. 8–18 c . Assuming either impending motion or motion of the belt, the magnitude of the frictional force dF = m dN. This force opposes the sliding motion of the belt, and so it will increase the magnitude of the tensile force acting in the belt by dT . Applying the two force equations of equilibrium, we have
R + F
x= 0; T cos a du
2 b + m dN - ( T + dT ) cos a du 2 b = 0 + QF
y= 0; dN - ( T + dT ) sin a du
2 b - T sina du 2 b = 0
Since du is of infinitesimal size , sin(du >2) = du>2 and cos(du>2) = 1 . Also, the product of the two infinitesimals dT and du >2 may be neglected when compared to infinitesimals of the first order. As a result, these two equations become
m dN = dT and
dN = T du Eliminating dN yields
dT
T = m du
Motion or impending motion of belt relative to surface
(a) r
T2
T1 b
u
(b)
T1 T2
u
dF m dN ds
(c) T dT
T y
dN
x du
2
du 2 du
2 du
2
Fig. 8–18
8
Integrating this equation between all the points of contact that the belt makes with the drum, and noting that T = T
1at u = 0 and T = T
2at u = b, yields
L
T2
T1
dT T = m
L
b 0
du
ln T
2T
1= mb Solving for T
2, we obtain
T
2= T
1e
mb(8–6)
where
T
2, T
1= belt tensions; T
1opposes the direction of motion (or impending motion) of the belt measured relative to the surface, while T
2acts in the direction of the relative belt motion (or impending motion); because of friction, T
27 T
1m = coefficient of static or kinetic friction between the belt and the surface of contact
b = angle of belt to surface contact, measured in radians e = 2.718 c, base of the natural logarithm
Note that T
2is independent of the radius of the drum, and instead it is a function of the angle of belt to surface contact, b. As a result, this equation is valid for flat belts passing over any curved contacting surface.
Motion or impending motion of belt relative to surface
(a) r
T2
T1 b
u
Fig. 8–18 (Repeated)
Flat or V-belts are often used to transmit the torque developed by a motor to a wheel attached to a pump, fan or blower.
8
The maximum tension that can be developed in the cord shown in
Fig. 8–19 a is 500 N. If the pulley at A is free to rotate and the coefficient of static friction at the fixed drums B and C is m
s= 0.25, determine the largest mass of the cylinder that can be lifted by the cord.
T
A
(a)
C B
D
45⬚ 45⬚
SOLUTION