Problem#1
An inventive child named Pat wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley (Fig. 1), Pat pulls on the loose end of the rope with such a force that the spring scale reads 250 N. Pat’s true weight is 320 N, and the chair weighs 160 N. (a) Draw free-body diagrams for Pat and the chair considered as separate systems, and another diagram for Pat and the chair considered as one system. (b) Show that the acceleration of the system is upward and find its magnitude. (c) Find the force Pat exerts on the chair. |
| Fig.1 |
(a) see figure to the right
 |
| Fig.2 |
T = 250 N in each rope. Applying
∑ F = ma
2T – 480 N = ma
where m = 480 N/(9.80 m/s2) = 49.0 kg
Solving for a gives
a = (500 N – 480 N)/(49.0 kg) = 0.408 m/s2
(c) ∑ F = ma = on Pat:
n + T – 320 N = ma
where m = 320 N/(9.80 m/s2) = 32.7 kg
then
n = (32.7 kg)(0.408 m/s2) + 320 N – 250 N = 83.3 N
Problem#2
A time-dependent force, F = (8.00i + 4.00tj) N, where t is in seconds, is exerted on a 2.00-kg object initially at rest. (a) At what time will the object be moving with a speed of 15.0 m/s? (b) How far is the object from its initial position when its speed is 15.0 m/s? (c) Through what total displacement has the object traveled at this time?
Answer:
(a) ∑F = ma gives the object’s acceleration
(8.00i + 4.00tj) N = (2.00 kg)a
a = (4.00i + 2.00tj) N/kg
acceleration can also be expressed as
a = dv/dt
or
v = ∫adt
v = (4.00ti + 1.00t2 j) m/s
(a) for |v| = 15.0 m/s, |v|2 = 225 m2/s2, so that
|v|2 = |vx|2 + |vy|2
225 m2/s2 = 16.0t2 m2/s4 + 1.00t4 m2/s6
t4 + 16.0s2t2 – 225 s4 = 0
(t2 – 9)(t2 + 25) = 0
t = 3 s
(c) Take ri = 0 at t = 0. The position is
r = ∫vdt = ∫0t (4.00ti + 1.00t2 j)dt
r = (2.00t2i + t3/3 j) m
at t = 3 s we evaluate.
r = (18.0i + 9.00j) m
(b) so
|r|2 = |x|2 + |y|2
|r|2 = (18.0)2 + (9.00)2
|r| = 20.1 m
Problem#3
To prevent a box from sliding down an inclined plane, student A pushes on the box in the direction parallel to the incline, just hard enough to hold the box stationary. In an identical situation student B pushes on the box horizontally. Regard as known the mass m of the box, the coefficient of static friction µs between box and incline, and the inclination angle θ. (a) Determine the force A has to exert. (b) Determine the force B has to exert. (c) If m = 2.00 kg, θ = 25.0°, and µs = 0.160, who has the easier job? (d) What if µs = 0.380? Whose job is easier?
Answer:
(a) Situation A
 |
| Fig.3 |
FA + fs – mg sin θ = 0
∑Fy = may
n – mg cos θ = 0
n = mg cos θ
so that
FA + µsmg cos θ – mg sin θ = 0
FA = mg(sin θ – µs cos θ)
(b) Situation B
 |
| Fig.4 |
FB cos θ + fs – mg sin θ = 0
∑Fy = may
– FB sin θ + n – mg cos θ = 0
Substitute n = mg cos θ + FB sin θ to find
FB cos θ + µs(mg cos θ + FB sin θ) – mg sin θ = 0
FB cos θ + µsmg cos θ + µsFB sin θ – mg sin θ = 0
FB (cos θ + µssin θ) = mg (sin θ – µs cos θ)
FB = mg (sin θ – µs cos θ)/(cos θ + µssin θ)
FB = FA/(cos θ + µssin θ)
(c) If m = 2.00 kg, θ = 25.0°, and µs = 0.160,
FA = (2.00 kg)(9.80 m/s2)(sin 25.00 – 0.160 cos 25.00)
FA = 5.44 N
And
FB = (5.44 N)/(cos 25.00 + 0.160 sin 25.00)
FB = 5.59 N
Student A need exert less force.
(d) FB = (5.44 N)/(cos 25.00 + 0.38 sin 25.00) = 5.08 N
Student B need exert less force.
Problem#4
Three blocks are in contact with each other on a frictionless, horizontal surface, as in Figure 5. A horizontal force F is applied to m1. Take m1 = 2.00 kg, m2 = 3.00 kg, m3 = 4.00 kg, and F = 18.0 N. Draw a separate free-body diagram for each block and find (a) the acceleration of the blocks, (b) the resultant force on each block, and (c) the magnitudes of the contact forces between the blocks. (d) You are working on a construction project. A coworker is nailing up plasterboard on one side of a light partition, and you are on the opposite side, providing “backing” by leaning against the wall with your back pushing on it. Every blow makes your back sting. The supervisor helps you to put a heavy block of wood between the wall and your back. Using the situation analyzed in parts (a), (b), and (c) as a model, explain how this works to make your job more comfortable.
 |
| Fig.5 |
(a)
Block m1: 18 N – P = (2 kg)a
Block m2 : P – Q = (3 kg)a
Block m3: Q = (4 kg)a
 |
| Fig.6 |
a = 2.00 m/s2
(b) Block m1: 18 N – P = (2 kg)(2.00 m/s2) = 4.00 N (net force on the 2 kg)
Block m2 : P – Q = (3 kg)(2.00 m/s2) = 6.00 N (net force on the 3 kg)
Block m3: Q = (4 kg)(2.00 m/s2) = 8.00 N (net force on the 4 kg)
(c) Block m1: 18 N – P = (2 kg)(2.00 m/s2) = 4.00 N
P = 18 N – 4 N = 14 N
Block m3: Q = (4 kg)(2.00 m/s2) = 8.00 N
(d) The 3-kg block models the heavy block of wood. The contact force on your back is represented by Q, which is much less than the force F. The difference between F and Q is the net force causing acceleration of the 5-kg pair of objects. The acceleration is real and nonzero, but lasts for so short a time that it never is associated with a large velocity. The frame of the building and your legs exert forces, small relative to the hammer blow, to bring the partition, block, and you to rest again over a time large relative to the hammer blow. This problem lends itself to interesting lecture demonstrations. One person can hold a lead brick in one hand while another hits the brick with a hammer.
Problem#5
An object of mass M is held in place by an applied force F and a pulley system as shown in Figure P5.55. The pulleys are massless and frictionless. Find (a) the tension in each section of rope, T1, T2, T3, T4, and T5 and (b) the magnitude of F. Suggestion: Draw a free-body diagram for each pulley.
 |
| Fig.7 |
(a) First, we note that F = T1. Next, we focus on the mass M and write
T5 = Mg
Next, we focus on the bottom pulley and write
T5 = T2 + T3
 |
| Fig.8 |
T4 = T1 + T2 + T3
Since the pulleys are not starting to rotate and are frictionless,
T1 = T3 and T2 = T3
From this information, we have
T5 = 2T2, so
T2 = Mg/2
Then
T1 = T2 = T3 = Mg/2
And
T4 = 3Mg/2
And
T5 = Mg
(b) Since F = T1, we have F = Mg/2.
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