Problem#1
One block of mass 5.00 kg sits on top of a second rectangular block of mass 15.0 kg, which in turn is on a horizontal table. The coefficients of friction between the two blocks are µs = 0.300 and µk = 0.100. The coefficients of friction between the lower block and the rough table are µs = 0.500 and µk = 0.400. You apply a constant horizontal force to the lower block, just large enough to make this block start sliding out from between the upper block and the table. (a) Draw a free-body diagram of each block, naming the forces on each. (b) Determine the magnitude of each force on each block at the instant when you have started pushing but motion has not yet started. In particular, what force must you apply? (c) Determine the acceleration you measure for each block.Answer:
(a), (b) Motion impending
fs2 = µn = 0.500 x 196 N = 98.0 N
then P = fs1 + fs2 = 14.7 N + 98.0 N = 112.7 N
(c) Once motion starts, kinetic friction acts.
112.7 N – 0.100(49.0 N) – 0.400(196 N) = (15.0 kg)a2
a2 = 1.96 m/s2
and
0.100(49.0 N) = (5.00 kg)a1
a1 = 0.980 m/s2
Problem#2
A 1.00-kg glider on a horizontal air track is pulled by a string at an angle !. The taut string runs over a pulley and is attached to a hanging object of mass 0.500 kg as in Fig. 2. (a) Show that the speed vx of the glider and the speed vy of the hanging object are related by vx = uvy, where u = z(z2 - h02)-1/2. (b) The glider is released from rest. Show that at that instant the acceleration a x of the glider and the acceleration ay of the hanging object are related by ax = uay. (c) Find the tension in the string at the instant the glider is released for h0 = 80.0 cm and θ = 30.0°.
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| Fig.2 |
(a) Let x represent the position of the glider along the air track. Then
z2 = x2 + h02
so that
x = (z2 – h02)1/2
then
vx = dx/dt = ½ (z2 – h02)-1/2(2z)dz/dt
Now dz/dt is the rate at which string passes over the pulley, so it is equal to vy of the counterweight.
vx = z(z2 – h02)-1/2vy = uvy
(b) ax = dvx/dt = duvy/dt = udvy/dt + vydu/dt
at release from rest, vy = 0 and
ax = udvy/dt = uay
(c) sin 30.00 = 80.0 cm/z
z = 1.60 cm
u = z(z2 – h02)-1/2
u = (1.6)(1.62 – 0.82)-1/2 = 1.15
then ax = 1.15ay
For the counterweight
∑Fy = may
T – (0.50 kg)(9.80 m/s2) = –(0.50 kg)ay
ay = –2T + 9.8
For the glider
∑Fx = max
T cos 30.00 = (1.00 kg)ax = 1.15ay
T cos 30.00 = 1.15(–2T + 9.8)
3.18 T = –2.31T + 11.3
T = 3.56 N
Problem#3
Cam mechanisms are used in many machines. For example, cams open and close the valves in your car engine to admit gasoline vapor to each cylinder and to allow the escape of exhaust. The principle is illustrated in Figure 3, showing a follower rod (also called a pushrod) of mass m resting on a wedge of mass M. The sliding wedge duplicates the function of a rotating eccentric disk on a camshaft in your car. Assume that there is no friction between the wedge and the base, between the pushrod and the wedge, or between the rod and the guide through which it slides. When the wedge is pushed to the left by the force F, the rod moves upward and does something, such as opening a valve. By varying the shape of the wedge, the motion of the follower rod could be made quite complex, but assume that the wedge makes a constant angle of θ = 15.0°. Suppose you want the wedge and the rod to start from rest and move with constant acceleration, with the rod moving upward 1.00 mm in 8.00 ms. Take m = 0.250 kg and M = 0.500 kg. What force F must be applied to the wedge?
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| Fig.3 |
(a) The upward acceleration of the rod is described by
yf = yi + vyit + ½ ayt2
10-3 m = 0 + 0 + ½ ay(8 x 10-3 s)2
ay = 31.2 m/s2
The distance y moved by the rod and the distance x moved by the wedge in the same time are related
by
tan 15.00 = y/x
x = y/tan 15.00
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| Fig.4 |
d2x/dt2 = (1/tan 15.00) d2y/dt2
ax = (1/tan 15.00)ay
ax = (1/tan 15.00)(31.2 m/s2) = 117 m/s2
The free body diagram for the rod is shown. Here H and H′ are forces exerted by the guide.
∑Fy = may
n cos θ – mg = may
n cos 15.00 – 0.250 kg(9.80 m/s2) = (0.250 kg)(31.2 m/s2)
n cos 15.00 = 10.3 N
n = 10.6 N
For the wedge,
∑Fx = max
F – n sin θ = Max
F – (10.6 N) sin 15.00 = (0.500 kg)(117 m/s2)
F = 58.3 N + 10.6 N sin 15.00
F = 61.1 N
Problem#4
Any device that allows you to increase the force you exert is a kind of machine. Some machines, such as the prybar or the inclined plane, are very simple. Some machines do not even look like machines. An example is the following: Your car is stuck in the mud, and you can’t pull hard enough to get it out. However, you have a long cable which you connect taut between your front bumper and the trunk of a stout tree. You now pull sideways on the cable at its midpoint, exerting a force f. Each half of the cable is displaced through a small angle θ from the straight line between the ends of the cable. (a) Deduce an expression for the force exerted on the car. (b) Evaluate the cable tension for the case where θ = 7.00° and f = 100 N.
Answer:
(a) Consider forces on the midpoint of the rope. It is nearly in equilibrium just before the car begins to move. Take the y-axis in the direction of the force you exert:
∑Fy = may
f – 2T sin θ = 0
T = f/2sinθ
(b) the cable tension for the case where θ = 7.00° and f = 100 N is
T = 100 N/(2 sin 7.000) = 410 N
Problem#5
Two blocks of mass 3.50 kg and 8.00 kg are connected by a massless string that passes over a frictionless pulley (Fig. 7). The inclines are frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string.
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| Fig.6 |
Since it has a larger mass, we expect the 8.00-kg block to move down the plane. The acceleration for both blocks should have the same magnitude since they are joined together by a non-stretching string. Define up the left hand plane as positive for the 3.50-kg object and down the right hand plane as positive for the 8.00-kg object.
(a) ∑Fx = mtotalax
m1g sin 35.00 – T + T – m2g sin 35.00 = (m1 + m2)a
(8.00 kg)(9.80 m/s2) sin 35.00 – (3.50 kg)(9.80 m/s2) sin 35.00 = (8.00 kg + 3.50 kg)a
a = 25.3 N/11.5 kg = 2.20 m/s2
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| Fig.7 |
m1g sin 35.00 – T = m1a
(8.00 kg)(9.80 m/s2) sin 35.00 – T = (8.00 kg)(2.20 m/s2)
44.97 N – T = 17.6 N
T = 27.4 N
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