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Chapter 11: Problem 24

In preparation for a demonstration on conservation of energy, a professor attaches a 5.00 -kg bowling ball to a 4.00 -m-long rope. He pulls the ball \(20.0^{\circ}\) away from the vertical and holds the ball while he discusses the physics principles involved. Assuming that the force he exerts on the ball is entirely in the horizontal direction, find the tension in the rope and the force the professor is exerting on the ball.

Short Answer

Expert verified
Answer: The tension in the rope is about 53.7 N, and the force the professor is exerting on the ball is about 18.3 N.

Step by step solution

01

Identify forces acting on the ball

First, we need to identify the forces acting on the ball. There are three forces acting on the ball: 1. Gravitational force (weight) acted downward by the ball: \(W = mg\), where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity (\(\approx 9.81 m/s^2\)). 2. Tension force in the rope: \(T\), which acts along the rope. 3. Horizontal force exerted by the professor: \(F_{p}\), which acts perpendicular to the rope.
02

Set up equilibrium equations

Since the ball is stationary, it is in equilibrium. Therefore, the net forces in both horizontal and vertical directions are equal to zero. We can write these two equations using the equilibrium conditions: 1. In the horizontal direction: \(T\sin(\theta) = F_{p}\) 2. In the vertical direction: \(T\cos(\theta) = mg\)
03

Calculate tension

We can start by solving the vertical equation for the tension in the rope: \(T\cos(\theta) = mg\) Now, we can plug in the known values: - \(m = 5.00\,\text{kg}\) - \(\theta = 20^{\circ}\) - \(g = 9.81\,\text{m/s}^2\) \(T = \frac{mg}{\cos(20^{\circ})} = \frac{(5.00 \,\text{kg})(9.81\,\text{m/s}^2)}{\cos(20^{\circ})}\) Calculating the tension, we get: \(T \approx 53.7 \,\text{N}\)
04

Calculate the horizontal force exerted by the professor

Now that we have the tension, we can use the horizontal equation to find the force exerted by the professor: \(F_{p} = T\sin(\theta)\) Plug in the known values: \(F_{p} = (53.7 \,\text{N})\sin(20^{\circ})\) Calculating the force, we get: \(F_{p} \approx 18.3 \,\text{N}\) Therefore, the tension in the rope is about \(53.7\,\text{N}\), and the force the professor is exerting on the ball is about \(18.3\,\text{N}\).

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