Solving Force on a Pulley: Calculating Ceiling Force

[pewpew23]

Homework Statement

A crate with a weight of W is being lifted with an upward acceleration equal to one-half the magnitude of acceleration due to gravity. The crate is being lifted by a rope that passes over a frictionless pulley connected at the ceiling as shown in the drawing. The mass of the rope and the pulley are negligible. What force does the ceiling apply to the pulley?

A) W
B) 3/2 W
C) $$\sqrt{}$$3/2 W
D) 3W
E) 4W

Homework Equations

F=ma

The Attempt at a Solution

the force the pulley applies on the ceiling is equal and opposite to the force the ceiling applies to the pulley. But just using F=ma, and plugging in (-9.79/2)(W) equals 4.9W
Am I looking at this problem wrong, or missing something? Do I need to include the force of gravity pulling down on the pulley somehow?

 

[jegues]

Do you know what the correct answer is by chance?

 

[pewpew23]

No, I don't have the correct answer.

 

[jegues]

I'm not entirely sure if this is correct but I'll throw it out there:

I'm going to write W in its "uglier" form for now (as simply mg).

F = ma so,

T - mg = ma, where a = g/2 so,

T = mg + (mg)/2

T = m(3/2g) since W = mg then,

T = 3/2W

 

[pewpew23]

ohhh... I had a feeling I had to include the full force of gravity in there. Thanks for your help!
I have a couple more questions left... is there any chance you could help me with those as well? I completely understand if that is asking for too much :)

 

[PhanthomJay]

You've got to draw free body diagrams. Isolate the crate and examine the forces, known and unknown, acting on it, then apply Newton 2 to solve for the unknown force. Note that the tension in the rope on each side of an ideal pulley are equal. Then isolate the pulley to solve for the ceiling force on it, using one of Newton's laws, and see what you get for an answer. NOTE: if you have more questions unrelated to this topic, please post them under a separate post.

 

[jegues]

If it's not part of this question then simply post another thread and I (along with several others) would be glad to help you.

 

[PhanthomJay]

I'm not entirely sure if this is correct but I'll throw it out there:

I'm going to write W in its "uglier" form for now (as simply mg).

F = ma so,

T - mg = ma, where a = g/2 so,

T = mg + (mg)/2

T = m(3/2g) since W = mg then,

T = 3/2W

That's not quite correct, T and the ceiling force are not the same. Without a drawing, I'm assuming that the rope makes a 180 degree around the pulley, is this correct?

 

[pewpew23]

http://i10.tinypic.com/2a66pap.jpg

It is similar to this, except with a block on one end and nothing on the other.

 

[jegues]

I'm trying to imagine what forces would be exerted on the pulley after we've figured out the tension in the rope. We know its accelerating upward but it still has to have some fraction of the weight of the box weighing down on the pulley.

Could you toss another hint perhaps?

 

[PhanthomJay]

I'm trying to imagine what forces would be exerted on the pulley after we've figured out the tension in the rope. We know its accelerating upward but it still has to have some fraction of the weight of the box weighing down on the pulley.

Could you toss another hint perhaps?

If the downward tension on one side of the pulley is 3W/2, then the tension pulling down on the other side of the pulley must also be 3W/2. Now draw the free body of the pulley and note the 2 forces acting downward, and the ceiling force acting upward, and note that the pulley isn't moving up or down, so now solve for the force of the ceiling on the pulley.

 

[jegues]

If the downward tension on one side of the pulley is 3W/2, then the tension pulling down on the other side of the pulley must also be 3W/2. Now draw the free body of the pulley and note the 2 forces acting downward, and the ceiling force acting upward, and note that the pulley isn't moving up or down, so now solve for the force of the ceiling on the pulley.

3W, thank you for clarifying that.