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

[Zhartek]

Homework Statement

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.

Homework 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:
$$\sum \vec{F} = 0$$
$$\sum \vec{\tau} = 0$$

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.

 

[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?

 

[Zhartek]

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?

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..

 

[BruceW]

Right, so by the definition of cross product, I get the third equation to be
$$RT + RF = 0$$

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!)