Solving Force on Pulley: 25kg Accelerating at 1.4m/s^2

[Jstew]

Homework Statement

A cart connected to a massless rope has mass 25kg. The rope passes over a fixed pulley (The rope is pulled upwards, the car moves to the right, the pulley is attached to a pole in the ground.) The cart accelerates horizontally at 1.4m/s^2, find the upward force on the pulley.

Homework Equations

F=ma

The Attempt at a Solution

The answer seems to either be F=(25kg)(1.4m/s^2) or F=0, but I can't decide which it is since I can't figure out what the free body diagram for the pulley would look like.

 

[Doc Al]

That does seem like an oddly-worded question, if I'm picturing the set up correctly. Are they asking for the required tension in the rope? Or the net force on the pulley? (Which we can assume is zero.) I would guess that they want the required upward force that you need to exert on the rope (which you have solved for).

 

[Jstew]

It seems very oddly-worded to me as well. I've copied the drawing as best I could and the wording of the problem verbatim. If the upward force on the pulley is non-zero it seems like it would lead to odd action-reaction pairs: The pulley exerts downward force F on the rope, the cart exerts downward force F on the rope, so T=2F. But shouldn't T=F in this problem?

 

[Doc Al]

If the upward force on the pulley is non-zero it seems like it would lead to odd action-reaction pairs: The pulley exerts downward force F on the rope, the cart exerts downward force F on the rope, so T=2F.

The cart exerts a horizontal force on the rope. (I don't understand your comment about "action-reaction" pairs. You have not identified any such pairs.)

But shouldn't T=F in this problem?

It does.

Since the rope (and pulley) is massless, it has a single tension throughout.

 

[Jstew]

To clarify the "action-reaction" comment, what exactly is exerting the force on the pulley and in which direction is this force?

 

[Doc Al]

I'd say that one section of the rope is exerting a horizontal force on the pulley, another section is exerting a vertical force on the pulley, and the pole supporting the pulley is exerting a third force on the pulley. (They add to zero, of course, since the pulley is not accelerating.)

 

[Jstew]

So then the pulley is exerting a horizontal force and a vertical force on the rope. (This is the action-reaction I was talking about) If the pulley is exerting force on the rope, how is T=ma and not T=ma+F where F is the force of the pulley on the rope.

 

[Doc Al]

So then the pulley is exerting a horizontal force and a vertical force on the rope.

Taking the rope as a whole, that is certainly true. Looking at the rope in sections: The pulley exerts a horizontal force (equal to T) on the horizontal section of rope, and a vertical force (also equal to T) on the vertical section of rope.

(This is the action-reaction I was talking about) If the pulley is exerting force on the rope, how is T=ma and not T=ma+F where F is the force of the pulley on the rope.

It looks like you are using "F" to stand for the entire force of the pulley on the rope taken as a whole. I'm not sure why you are doing that. In any case, what horizontal forces act on the cart? I only see one: the tension in the rope, T. Thus T = ma. (The force of the pulley on the entire rope is irrelevant.)