How Do You Calculate Tension and Pole Force on a Leaning Tether Ball?

[Edwardo_Elric]

Homework Statement

A tether ball leans against the post to which it is attached. If the string to which the ball is attached is 1.80m long, the ball has a radius of .200m, and the ball has a mass of .500kg, what are the tension in the rope and the force the pole exerts on the ball? assume that there is so little friction between the ball and the pole that it can be neglected. ( The string is attached to the ball such that a line along the string passes through the center of the ball.)

Homework Equations

$$F_{x} = 0$$
$$F_{y} = 0$$

The Attempt at a Solution

Part I
Required: tension on the rope
T = ??
wball = (.500kg)(9.8m/s^2) = 4.9N
$$\sum{F_{y}} = 0 = T\cos{\theta} + (-w_{ball})$$
$$T = \frac{4.9N}{\cos{\theta}}$$
*I computed theta using its distances:
sin(theta) = x_{radius} / h_{tension}
= arcsin {0.200m/1.80m}
= 6.38 degress
$$T = \frac{4.9N}{\cos{6.38}}$$

T =4.93N

Part II
reqd: force on pole on ball:
F_{pole on ball} = ?
\sum{F_{x}} = 0 = T\sin{\theta} - F_{p.on.B}
(4.93N)sin{6.38} = F_{pole on ball}
F = .548N?

 

[andrevdh]

You've got both answers correct.

 

[m2003]

Your solution for Part I, finding the tension in the rope, is correct. However, your solution for Part II, finding the force the pole exerts on the ball, is incorrect. The correct solution for Part II is as follows:

First, we need to find the horizontal component of the tension force, which is equal to Tsinθ. This force is acting in the opposite direction to the force exerted by the pole on the ball. Therefore, we can set up the following equation:

Tsinθ = Fpole on ball

Substituting in the values we know, we get:

(4.93N)sin6.38° = Fpole on ball

Fpole on ball = 0.56N

This is the force that the pole exerts on the ball. This force is necessary to keep the ball in equilibrium, as it balances out the horizontal component of the tension force.