Question

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

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

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.

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: \(T\cos(\theta) = mg\)

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: \(T \approx 53.7 \,\text{N}\)

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 \(18.3\,\text{N}\).

Key concepts

Tension in Rope

When tackling a problem involving tension, imagine a rope as not just a simple string, but as an object that can exert forces. Here, the tension is what keeps the bowling ball suspended without falling. The tension force is directed along the rope, opposing the weight of the ball.

Tension, a restoring force, acts effectively to counterbalance any weight acting on the rope. It's often calculated in scenarios involving objects suspended or connected by ropes or strings.

To calculate the tension in this exercise, we use the component of tension force acting vertically. This accounts for the pull that counters the gravitational force and holds the ball still in its suspended position. By using trigonometry, tension is pulled horizontally, forming a component. The equation
\[T\cos(\theta) = mg\]
is used to solve for tension since the vertical force is balanced by gravity. By manipulating this equation, the tension can be extracted as:
\[T = \frac{mg}{\cos(\theta)}\]

This formula considers the mass of the object and the angle from the vertical to ensure balance in forces.

Gravitational Force

Gravitational force is a fundamental interaction that causes mutual attraction between objects with mass. For the bowling ball in the exercise, gravitational force is what pulls it downward towards the Earth.

Calculating this force is straightforward using the equation:
\[W = mg\]
where \(W\) is the weight (gravitational force), \(m\) is the mass, and \(g\) is the acceleration due to gravity, approximately 9.81 \(m/s^2\) on Earth.

In the problem, the professor needs to consider this force to maintain the equilibrium of the bowling ball. It is one of the three forces in balance with the tension in the rope and the horizontal force exerted by the professor. The gravitational force remains constant regardless of the rope's angle or the professor's exertion, highlighting the necessity of considering such a non-variable in physics problems.

Gravitational force plays a crucial role in problems involving balance and equilibrium, especially when an object, like a ball, is suspended.

Equilibrium Equations

In physics, equilibrium signifies a state where all forces acting on an object are balanced. This indicates that the net force is zero, resulting in no movement. This concept is vital in determining how forces like tension or gravitational force interplay with each other.

In the exercise addressed, equilibrium equations make it possible to understand how the ball is stationary despite multiple forces acting upon it. The conditions for equilibrium are:

1. The sum of horizontal forces equals zero.
2. The sum of vertical forces equals zero.

For our scenario, the equilibrium equations can be defined as:

**Horizontal Equation:**
\[T\sin(\theta) = F_{p}\]

**Vertical Equation:**
\[T\cos(\theta) = mg\]

Each part of these equations maintains the ball's stationary state by ensuring that the tension in the rope and the forces applied by the professor balance out the weight of the bowling ball. These equations are perfect for solving scenarios involving ropes or strings, particularly when combined forces are presented in multiple directions.