Applications of Newton’s Laws Problems and Solutions

 Problem#1

A 3.00-kg object is moving in a plane, with its x and y coordinates given by x = 5t² – 1 and y = 3t³ + 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00 s.

Answer:

Given: x = 5t² – 1 and y = 3t³ + 2

the speed function is obtained from

v_x = dx/dt = 10t and v_y = dy/dt = 9t²

the acceleration function is obtained from

a_x = dv_x/dt = 10 m/s² and a_y = dv_y/dt = 18t

at t = 2.00 s, a_x = 10 m/s² and a_y = 18 x 2.00 s = 36.0 m/s²

ΣFx = ma_x : 3.00 kg(10.0 m/s²) = 30.0 N
ΣFy = ma_y : 3.00 kg(36.0 m/s²) = 108 N

Then
ΣF² = ΣF_x² + ΣF_y²

ΣF = √[(30.0 N)² + (108 N)²] = 112 N

Problem#2
The distance between two telephone poles is 50.0 m. When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.200 m. Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.

Answer:

Fig.1

Given: m = 1.00 kg; mg = 9.80 N  θ

tan θ = 0.200 m/25.0 m
θ = 0.458⁰

Balance forces,

2T sin θ = mg

T = mg/2sinθ = 9.80 N/(2 sin 0.458⁰) = 613 N


Problem#3
A bag of cement of weight 325 N hangs from three wires as suggested in Figure 2. Two of the wires make angles θ₁ = 60.0° and θ₂ = 25.0° with the horizontal. If the system is in equilibrium, find the tensions T₁, T₂, and T₃ in the wires.

Fig.2

Answer:
From figure, we get

T₃ = mg                                                                 (1)

T₁ sin θ₁ + T₂ sin θ₂ = mg                                (2)

T₁ cos θ₁ = T₂ cos θ₂                                         (3)

Fig.3

multiply equation 1 with cos θ₂ and equation 2 with sin θ₂ then eliminate T₂ and solve for T₁

T₁ sin θ₁ cos θ₂ + T₂ sin θ₂ cos θ₂ = mg cos θ₂          (4)
             
T₁ cos θ₁ sin θ₁ = T₂ cos θ₂ sin θ₁                                 (5)

add up the equations 4 and 5 we get

T₁(sin θ₁ cos θ₂ + cos θ₁ sin θ₂) = mg cos θ₂

T₁ = mg cos θ₂/sin (θ₁ + θ₂)

So,

T₁ = 325 N cos 25.0⁰/sin (60.0⁰ + 25.0⁰) = 296 N

T₃ = mg = 325 N

T₂ = T₁(cos θ₁/cos θ₂) = 296 N (cos 60.0⁰/cos 25.0⁰) = 163 N

Problem#4
A bag of cement of weight mg hangs from three wires as shown in Figure 2. Two of the wires make angles θ₁ and θ₂ with the horizontal. If the system is in equilibrium, show that the tension in the left-hand wire is

T₁ = F_g cos θ₂/sin (θ₁ + θ₂)

Answer:
See the solution for T₁ in Problem 3.

Problem#5
You are a judge in a children’s kite-flying contest, and two children will win prizes for the kites that pull most strongly and least strongly on their strings. To measure string tensions, you borrow a weight hanger, some slotted weights, and a protractor from your physics teacher, and use the following protocol, illustrated in Figure 4: Wait for a child to get her kite well controlled, hook the hanger onto the kite string about 30 cm from her hand, pile on weight until that section of string is horizontal, record the mass required, and record the angle between the horizontal and the string running up to the kite. (a) Explain how this method works. As you construct your explanation, imagine that the children’s parents ask you about your method, that they might make false assumptions about your ability without concrete evidence, and that your explanation is an opportunity to give them confidence in your evaluation technique. (b) Find the string tension if the mass is 132 g and the angle of the kite string is 46.3°.

Fig.4

Answer:
(a) An explanation proceeding from fundamental physical principles will be best for the parents and for you. Consider forces on the bit of string touching the weight hanger as shown in the free-body diagram:

Horizontal Forces: ΣF_x = max : −T_x + T cos θ =0

Vertical Forces: ΣF_y = ma_y : −mg + T sin θ =0

You need only the equation for the vertical forces to find that the tension in the string is given by T = mg/sinθ. The force the child feels gets smaller, changing from T to T cos θ , while the counterweight hangs on the string. On the other hand, the kite does not notice what you are doing and the tension in the main part of the string stays constant. You do not need a level, since you learned in physics lab to sight to a horizontal line in a building. Share with the parents your estimate of the experimental uncertainty, which you make by thinking critically about the measurement, by repeating trials, practicing in advance and looking for variations and improvements in technique, including using other observers. You will then be glad to have the parents themselves repeat your measurements.

(b) T = mg/sin θ = (0.132 kg)(9.80 m/s²)/sin 46.3⁰ = 1.79 N   

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