# Homework Help: Determining Tension in a rope

1. Oct 16, 2014

### Ali0086

An m = 12.4kg mass is attached to a cord that is wrapped around a wheel of radius r = 10.0cm (see the figure below). The acceleration of the mass down the frictionless incline is measured to be a = 2.10m/s^2.
a) Assuming the axle of the wheel to be frictionless, and the angle to be θ = 36.2deg determine the tension in the rope.
b) Determine the moment of inertia of the wheel
c)Determine angular speed of the wheel 2.04s after it begins rotating, starting from rest.

All in all I just have a lot of trouble drawing free body diagrams and just figuring out how to approach a problem with my given set of known variables.

On my initial attempt for part a) I tried drawing a free body diagram but I didn't know if I should add the value I get from mgsin(theta) to the value that I got for F (I used m*a to get that value).

I've uploaded my attempt at a solution and I didn't get very far. I feel like there's a process at solving these questions but I just don't get it. Any help would be appreciated.

2. Oct 16, 2014

### Dr.D

Why do you think that there is a force F? It is not mentioned in the problem statement.

I think you are a victim of Mr. d'Alembert!!

Draw your FBDs with on the forces (the actual forces, not those you imagine) acting on the bodies. Then sum forces and set Sum F = m*a.

3. Oct 16, 2014

### Ali0086

So there is no force at all acting on the block? If they give you an acceleration and a mass, wouldnt the product of those two be the force heading down the ramp? That's something that confuses me in FBDs. When is there actually a force present? When one is being constantly applied by an external object only?

4. Oct 16, 2014

### Dr.D

No one said that there are no forces on the block. It is pretty clear that there are forces (1) normal to the ramp and acting on the block, ( 2) a force (tension) in the cord. If friction were involved, this would be yet another force. Oh, and don't forget the force of gravity (the weight). All of these are real forces; m*a is NEVER a force.

Read what Newton's law says: Sum F = m*a
All of the (real) forces go on the left side.

5. Oct 16, 2014

### Dr.D

One more thing: For our purposes here, forces are either (1) contact forces, applied by a push or a pull through direct contact, or they are field forces, such as gravity. The first group, the contact forces, require that something be "attached" to the object to apply the force, such as a rope, a floor, or a hand, all of which are "in contact" with the object. Gravity, on the other hand, acts at a distance, without any direct agent to apply it. Thus a satellite in earth orbit is subject to earth's gravity, even though there is no agent (a rope, a hand, etc) to apply it.

6. Oct 16, 2014

### Ali0086

Alright, so i got these two equations now, am I on the right track or did I go wrong somewhere?

ΣFx = m*a = mgSinθ - T
Σtorque = I*α = ½Mr^2*α= ½Mr^2*(a/r)

edit: Nvm I got the correct answer

I used ΣFx = m*a = mgSinθ - T to find my tension and worked on from there

Thanks a lot for your help!!

Last edited: Oct 16, 2014