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!)

 

[asteriatic]

I would first commend the effort put into finding equations and attempting to solve the problem. However, I would also suggest double checking the equations and the forces acting on the ball. It seems that there may be an error in the third equation, as the torque from the string and the friction force should not be included in the torque equation. Additionally, the normal force should be perpendicular to the plane and the friction force should be parallel to the plane, so the angles in the equations may need to be adjusted accordingly.

Once the equations are corrected, I would suggest plugging in the given values and solving for the unknowns. This problem can also be solved using a free body diagram, where the forces acting on the ball are drawn and labeled, and then the equations of equilibrium can be applied.

In terms of the physical interpretation of the problem, it is important to consider the forces and torques acting on the ball and how they balance out to keep the ball in place at the given angle. This type of problem is commonly encountered in mechanics and can be applied to real-world scenarios, such as objects held in place on an inclined plane or hanging from a string.

Overall, it is important to carefully analyze and understand the problem and the forces and torques involved in order to correctly solve it. Keep up the good work and keep practicing problem-solving skills in physics!