Compound pulley tension question

[sailsinthesun]

[SOLVED] Compound pulley tension question

Homework Statement

An object of mass M is held in place by an applied force and a pulley system as shown in Figure P5.55. The pulleys are massless and frictionless.
http://img520.imageshack.us/img520/7840/p445fw5.th.gif
(a) Find the tension in each section of rope, T1, T2, T3, T4, andT5. (Answer in terms of M and g.)
(b) Find the magnitude of F.

The Attempt at a Solution

The only thing I know is that T5=Mg. I'm trying to draw the a FBD for each pulley but I'm getting confused with relating T1, T2, T3, and T4 with Mg and F.

 

[Doc Al]

The only thing I know is that T5=Mg. I'm trying to draw the a FBD for each pulley but I'm getting confused with relating T1, T2, T3, and T4 with Mg and F.

It's easier than it looks. The FBD for each pulley should be easy since the only forces on the pulleys are the string tensions.
Hint 1: The tension is the same throughout a continuous piece of rope (since the pulleys are massless and frictionless).
Hint 2: How does F relate to T1?

 

[sailsinthesun]

It's easier than it looks. The FBD for each pulley should be easy since the only forces on the pulleys are the string tensions.
Hint 1: The tension is the same throughout a continuous piece of rope (since the pulleys are massless and frictionless).
Hint 2: How does F relate to T1?

F-T1=0 so F=T1?

When I'm drawing the FBD for the top pulley, what would the downward force be? F?

 

[Doc Al]

F-T1=0 so F=T1?

Right!

When I'm drawing the FBD for the top pulley, what would the downward force be? F?

The downward force will equal the three tensions that pull on it.

 

[sailsinthesun]

Right!


The downward force will equal the three tensions that pull on it.

So the downward force would be T1+T2+T3? Meaning T4=T1+T2+T3? And how do I relate these tensions to M and g?

 

[Doc Al]

So the downward force would be T1+T2+T3? Meaning T4=T1+T2+T3?

Right.

And how do I relate these tensions to M and g?

By combining that with other equations you can come up with. What about the second pulley? And, especially, what about my hint #1?

 

[sailsinthesun]

Right.

By combining that with other equations you can come up with. What about the second pulley? And, especially, what about my hint #1?

Even if the rope goes through a pulley? So that means T1=T2=T3?

Ahh, just solved it. Thanks for all of your help. T1=T2=T3=Mg/2 and since T4=T1+T2+T3, T4=(3Mg)/2 and since F=T1, F=Mg/2

 

[JinFTW]

Hey guys I have a problem I got wrong similar to this. The pulley set up is exactly the same, but the outside rope is pulled at an angle (60) and I was curious to see how the force changes with the angle. Don't get me wrong here, I know that to find the x component of the force it would be FCos(60) and the find the y component you would use FSin(60), but do you use 1/2MG as F even though its pulled at an angle? And furthermore, do you do you square Fx and Fy and then take the root of it to find the final F?