In preparation for a demonstration on conservation of energy, a professor attaches a 8.7-kg bowling ball to a 3.2-m-long rope. He pulls the ball 18° 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.

WEll, he has to counter the horizontal force on the ball.

draw the force diagram mg down,horizontal, and tension
costheta=mg/Tension

figure the cosine of 18 degrees, solve for tension.

To find the tension in the rope, we can start by analyzing the forces acting on the bowling ball.

The main forces involved are:
1. The weight force acting downwards due to gravity.
2. The tension force in the rope acting towards the center of the circular path, which prevents the ball from falling straight down.
3. The force exerted by the professor in the horizontal direction.

Since the ball is in equilibrium and not accelerating, the net force acting on it in the horizontal direction must be zero. This means that the horizontal component of the tension force and the horizontal component of the force exerted by the professor must equal each other.

First, let's find the horizontal component of the tension force:
Tension_horizontal = Tension * cos(θ)

Where θ is the angle the rope makes with the vertical.

Next, let's find the horizontal component of the force exerted by the professor:
Force_horizontal = Force * cos(θ)

Since the net horizontal force is zero:
Tension_horizontal = Force_horizontal

Now we have enough information to find the tension in the rope.

1. Calculate the horizontal component of the tension force:
Tension_horizontal = Tension * cos(18°)

2. Calculate the horizontal component of the force exerted by the professor:
Force_horizontal = Force * cos(18°)

Since Tension_horizontal and Force_horizontal are equal, we have:
Tension * cos(18°) = Force * cos(18°)

Cancelling cos(18°) from both sides of the equation:
Tension = Force

Therefore, the tension in the rope is equal to the horizontal force exerted by the professor.

To find the tension, we need to know the value of the horizontal force exerted by the professor. This information is not provided in the question, so we cannot find the actual numerical value of the tension in the rope without that additional information.