In preparation for a demonstration on conservation of energy, a professor attaches a 7.2-kg bowling ball to a 3.4-m-long rope. He pulls the ball 23° away from the vertical and holds the ball while he discusses the physics principles involved. Assume that the force he exerts on the ball is entirely in the horizontal direction. Find the tension in the rope.
Find the force the professor is exerting on the ball.
To find the force the professor is exerting on the ball, we can break down the problem into its components.
First, let's analyze the forces acting on the bowling ball when it is at an angle of 23° from the vertical.
There are two forces acting on the ball: the tension force in the rope (T) and the gravitational force (mg), where m is the mass of the ball and g is the acceleration due to gravity.
Given:
Mass of the ball (m) = 7.2 kg
Angle from the vertical (θ) = 23°
Length of the rope (L) = 3.4 m
Acceleration due to gravity (g) = 9.8 m/s^2
To find the gravitational force acting on the ball, we can calculate the weight of the ball using the formula:
Weight = mass × gravity
Weight = (7.2 kg) × (9.8 m/s^2)
Weight = 70.56 N
Since the ball is not accelerating vertically, the tension in the rope (T) must be equal in magnitude to the gravitational force (mg), but in the opposite direction.
T = -mg
T = -70.56 N
However, this is only the vertical component of the tension force. We need to find the horizontal component of the tension force.
To find this component, we can use trigonometry. The horizontal component of the tension force (T_h) can be found using the relationship:
T_h = T × cos(θ)
T_h = (-70.56 N) × cos(23°)
T_h ≈ -64.8 N
Note that we obtained a negative sign because we are considering the tension force in the opposite direction to the gravitational force.
Therefore, the force the professor is exerting on the ball is approximately 64.8 Newtons in the horizontal direction.