# Homework Help: A ball tied up at an angular plane: Forces and torque

1. Feb 27, 2012

### Zhartek

1. The problem statement, all variables and given/known data
A spherical ball with radius R and mass M is kept in place at a plane with angle θ. The ball is kept in place with a string, as shown in this picture:

Given R = 20cm, M = 3.0kg, and θ = 30°, find:
1) the tension of the string.
2) the normal force on the ball from the plane.
3) the size of the friction force on the ball.

2. Relevant equations

I assume the ball to be in static equlibrium, so the sum of forces acting on the ball must be zero, as must the sum of torques:
$$\sum \vec{F} = 0$$
$$\sum \vec{\tau} = 0$$

3. The attempt at a solution
Based on the information given, I've found 4 forces acting on the ball: The gravity G, the normal force N, the frictional force F, and the force from the string T.

Based on this, I've come up with 3 equations:
$$\vec{T} + \vec{F_x} = \vec{N_x} \rightarrow T\hat{x} + Fcos(\theta)\hat{x} = N cos(90-\theta)\hat{x}$$
$$\vec{N_y} + \vec{F_y} = M\vec{g} \rightarrow Nsin(90-\theta)\hat{y} + Fsin(\theta)\hat{y} = Mg\hat{y}$$
$$\vec{T} \times \vec{R} + \vec{F} \times \vec{R} + \vec{N} \times \vec{R} = 0$$

My real problem is, I can't figure anything out from the third equation. Is my problem purely mathematical? Or have I made a mistake during the decomposing of the forces?

Thanks in advance for any advice! Also sorry for any (big) spelling mistakes.

2. Feb 27, 2012

### BruceW

You have decomposed the forces correctly, but you can do more with the third equation. Have a look at each of the terms, and look at the (very neat) diagram you have drawn. Can you immediately evaluate one of the terms?

3. Feb 28, 2012

### Zhartek

Right, so by the definition of cross product, I get the third equation to be
$$RT + RF = 0$$
Solving this system of equations yields
$$T = -109.7$$
$$F = 109.7$$
$$N = -29.4$$
Shouldn't N be positive? And the size of T and F seems a little big to me..

4. Feb 28, 2012

### BruceW

This isn't quite right, which is why the rest of the calculation will have gone awry. You had the equation:
$$\vec{T} \times \vec{R} + \vec{F} \times \vec{R} + \vec{N} \times \vec{R} = 0$$
And each of the R vectors correspond to the position vectors where each of the forces are applied. In other words, the R vectors are generally different for each. And using the definition of the cross product, we don't get RT + RF = 0 (but you are close!)