Problem#1
Two objects are connected by a light string that passes over a frictionless pulley, as in Figure 1. Draw free-body diagrams of both objects. If the incline is frictionless and if m1 = 2.00 kg, m2 = 6.00 kg, and θ = 55.0°, find (a) the accelerations of the objects, (b) the tension in the string, and (c) the speed of each object 2.00 s after being released from rest. |
| Fig.1 |
Answer:
Given: m1 = 2.00 kg, m2 = 6.00 kg, and θ = 55.00,
(a) the accelerations of the objects,
∑Fx = max
m2g sin θ – T = m2a (1)
and
∑Fy = may
T – m1g = m1a
From eq. (1) and (2), we get
a = (m2g sin θ – m1g)/(m1 + m2)
a = (6.00 kg sin 55.00 – 2.00 kg)(9.80 m/s2)/(2.00 kg + 6.00 kg)
a = 3.57 m/s2
T – m1g = m1a
T – 2.00 kg x 9.80 m/s2 = 2.00 kg x 3.57 m/s2
T = 26.7 N
(c) the speed of each object 2.00 s after being released from rest. (Given vi = 0), then
vf = at = 3.57 m/s2 x 2 s = 7.14 m/s
Problem#2
A tow truck pulls a car that is stuck in the mud, with a force of 2 500 N as in Fig. 2. The tow cable is under tension and therefore pulls downward and to the left on the pin at its upper end. The light pin is held in equilibrium by forces exerted by the two bars A and B. Each bar is a strut: that is, each is a bar whose weight is small compared to the forces it exerts, and which exerts forces only through hinge pins at its ends. Each strut exerts a force directed parallel to its length. Determine the force of tension or compression in each strut. Proceed as follows: Make a guess as to which way (pushing or pulling) each force acts on the top pin. Draw a free-body diagram of the pin. Use the condition for equilibrium of the pin to translate the free-body diagram into equations. From the equations calculate the forces exerted by struts A and B. If you obtain a positive answer, you correctly guessed the direction of the force. A negative answer means the direction should be reversed, but the absolute value correctly gives the magnitude of the force. If a strut pulls on a pin, it is in tension. If it pushes, the strut is in compression. Identify whether each strut is in tension or in compression.
 |
| Fig.2 |
We assume the vertical bar is in compression, pushing up on the pin with force A, and the tilted bar is in tension, exerting force B on the pin at -50.
Resultant force on the x-axis,
∑Fx = 0
–2500 N cos 300 + B cos 50.00 = 0
B = (2500 N cos 300)/cos 50.00 = 3370 N
–2500 N sin 300 + A – B sin 50.00 = 0
–2500 N sin 300 + A – (3370 N) sin 50.00 = 0
A = 3830 N
Positive answers confirm that
B is in tension and A is in compression.
Problem#3
Two objects with masses of 3.00 kg and 5.00 kg are connected by a light string that passes over a light frictionless pulley to form an Atwood machine, as in Figure 3. Determine (a) the tension in the string, (b) the acceleration of each object, and (c) the distance each object will move in the first second of motion if they start from rest.
 |
| Fig.3 |
Answer;
First, consider the 3.00 kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes
∑Fy = ma
T – 29.4 N = (3.00 kg)a (1)
The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction down for this mass, we have
∑Fy = ma
49 N – T = (5.00 kg)a (2)
Equations (1) and (2) can be solved simultaneously by adding them:
T – 29.4 N + 49.0 N – T = (3.00 kg)a + (5.00 kg)a
(b) This gives the acceleration as
a = 19.6 N/8.00 kg = 2.45 m/s2
(a) Then
T – 29.4 N = (3.00 kg)(2.45 m/s2) = 7.35 N
The tension is
T = 36.8 N
(c) Consider either mass. We have
yf = yi + vit + ½ at2
yf = 0 + 0 + ½ (2.45 m/s2)(1.00 s)2 = 1.23 m
Problem#4
In Figure 4, the man and the platform together weigh 950 N. The pulley can be modeled as frictionless. Determine how hard the man has to pull on the rope to lift himself steadily upward above the ground. (Or is it impossible? If so, explain why.)
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| Fig.4 |
Anwer:
As the man rises steadily the pulley turns steadily and the tension in the rope is the same on both sides of the pulley. Choose man-pulleyand-platform as the system:
∑Fy = ma
+T – 950 N = 0
T = 950 N
The worker must pull on the rope with force 950 N
Problem#5
In the Atwood machine shown in Figure 3, m1 = 2.00 kg and m2 = 7.00 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.40 m/s downward. (a) How far will m1 descend below its initial level? (b) Find the velocity of m1 after 1.80 seconds.
Answer:
Both blocks move with acceleration
a = (m2 – m1)g/(m1 + m2)
a = (7 kg – 2 kg)(9.80 m/s2)/(2 kg + 7 kg) = 5.44 m/s2
(a) Take the upward direction as positive for m1.
vxf = vxi + at
vxf = –2.40 m/s + (5.44 m/s2)(1.80 s) = 7.40 m/s (upward)
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