A tether ball leans against the post to which it is attached?

Question

A tether ball leans against the post to which it is attached. The string is attached to the ball such that a line along the string passes through the center of the ball. The string to which the ball is attached has length 1.50 and the ball's radius is 0.111 and has a mass of 0.271 . Neglect any friction between the ball and the pole.

A.What is the tension in the rope?
Take the free fall acceleration to be = 9.80 m/s^2

B.What is the force the pole exerts on the ball?
Take the free fall acceleration to be = 9.80 m/s^2

I found Part A to be 2.66N, but i can't figure out B

Answer 1

sin α = r/(r+l)

α = arcsin(r/(r+l))

A)
T cos α = mg,
T = mg sec α = mg(r+l) / √(l(l +2r))

B)
F = T sin α = mg tan α = mgr / √(l(l +2r))
F = 0.1841 N

Answer 2

(a) Tension in the rope:

First find the angle. Draw the triangle where the verticle leg of the triangle is the pole and the horizontal leg of the triangle is the radius of the ball. The length of the hypoteneuse will be the rope length + radius of the ball. You have to draw the triangle this way because the force is going to act on the center of the ball, so you have to draw the triangle where one of the vertices is the center of the ball. Anyway, now you can solve for the angle between the rope and the pole:

theta=arcsin[0.111/(0.111+1.5)]=0.0689 radians=3.95 degrees

You now find the tension in the rope:

T*cos(theta)=m*g=2.656N

Therefore, T=2.662N

(b) Force the pole exerts on the ball.

The pole only exerts a force on the ball where the ball touches the pole. Since the rope is pulling on the ball at an angle, there is a component of the rope tension that is pushing the ball toward the pole. However, the ball isn't moving, so you know there is a force that the pole exerts on the ball that is exactly equal and opposite to the horizontal component of the tension on the rope. So to find the magnitude of the force the pole is exerting on the ball, you find the horizontal component of the tension on the rope.

F=T*sin(theta)

You know T. You know theta. Compute F to find 0.183N horizontal force.

Now double check yourself. You have the horizontal component of the force in this triangle. You have the vertical component of the force in this triangle. You also have the diagonal component. Do the vector sums add up correctly? If they do, you are done. If not, you did something wrong.

Answer 3

I will suppose the dimensions are in meters. The horizontal force on the ball will be proportional to the sine of the angle between the vertical pole and the inclined string. Since the ball has a radius of 0.111/2 = 0.0555 m and the string is 1.5 m the sine will be 0.0555 /1.5 = 0.0037.
The horizontal force will be 0.0037 x 2.66 N or 0.0098 N
(As an aside, for small angles either the sine of the tangent can be used because they are closely the same.)

Answer 4

the rope is acting at a slight angle... to find it, draw a right-triangle with the hypotenuse 1.611m and another side 0.111m. Use 1.611 because the ball will touch the pole at it's center, so it will add length to the rope.

draw a fbd and you get mg = T cos(theta) where theta is the angle between the rope and the pole
solve for T and you get the Tension in the rope

The force of the pole on the ball is T sin(theta)

I forgot my calculator so I can't do sin(0.111/1.611), but that is your theta

Answer 5

Fp = T*(.111/1.50) where (as STATED), .111 is the ball's RADIUS.