Simple Pulley Question: Solving for Force on a Frictionless Pulley

[sjeddie]

Homework Statement

http://img70.imageshack.us/img70/3438/physicsquestiondiagram.png
A man sitting in a cart is attached by a rope of negligible mass that runs over a frictionless pulley. The rope falls in front of the man, so the man tries to pull on the rope in order to pull himself up. The total mass of the man and the cart is 100kg. What force should the man exert on the rope to make himself move up at a constant velocity?

Homework Equations

F=ma

The Attempt at a Solution

I did:
Let F be the force of the man pulling, T be the tension of the rope
then,
F - T = 0 and 100*9.8 - T = 0
so F = 980N

but, the solution tells me that,
2T = mg = 100*9.8
T = 490N
F = 490N

it doesn't make sense to me.. help?

 

[Doc Al]

Let F be the force of the man pulling, T be the tension of the rope
then,
F - T = 0

This is correct.

and 100*9.8 - T = 0

But this is not.

Hint: How many times does the rope pull up on the man and cart?

 

[sjeddie]

Once? I think.. Because there would be a tension force pulling up on the the cart and man, and while that's happening the man pulling down the rope would cancel with the other tension force.

What I said must have been wrong. please correct me.

 

[Doc Al]

Because there would be a tension force pulling up on the the cart and man, and while that's happening the man pulling down the rope would cancel with the other tension force.

How can the force exerted by the man on the rope cancel the force of the rope on the man? Forces can only cancel if they act on the same object.

Picture an imaginary box surrounding the "man + cart" system. How many strands of rope are attached to it and thus pulling up on it? How much force does strand exert?

 

[sjeddie]

Do you mean something like this?
http://img63.imageshack.us/img63/5387/pic1p.png
I don't understand how there can be 2 tension forces pulling him up. I mean, 1 is pulling on the box, would the other be pulling on his hand? but isn't his hand pulling down the rope and therefore contributing to that first tension force? Something like this
http://img129.imageshack.us/img129/3417/pic2h.png

 

[Doc Al]

Do you mean something like this?
http://img63.imageshack.us/img63/5387/pic1p.png

[/URL]
Yes, something like that. Except for Fa--what's that stand for? There are only three forces acting: The two tensions from the rope pulling up and gravity pulling down.

I don't understand how there can be 2 tension forces pulling him up. I mean, 1 is pulling on the box, would the other be pulling on his hand?

Sure. His hand (and him) are part of the same "man + cart" system.

but isn't his hand pulling down the rope and therefore contributing to that first tension force?

Absolutely. The force that he exerts on the rope equals the tension in the rope. So?

Something like this
http://img129.imageshack.us/img129/3417/pic2h.png

[/URL]
I don't understand this picture. What's Fa? (Don't forget that the cart and the man exert forces on each other. We don't have to worry about those internal forces if we just consider the "man + cart" as a single system.)

Compare these two situations:
(1) You tie a single rope to a tree limb and hang from it. If you weigh 150 lbs, what's the tension in the rope?
(2) You loop a rope over the tree limb, grab both ends and hang from it. What's the tension in the rope now? What if, instead of holding one end of the rope in each hand, you tie one end to your belt? Does that change anything?

 

[sjeddie]

Ohhhh! I get it now, I was considering the man pulling down on the rope as an external force instead of an internal force. Now it all makes sense. Thank you very much Doc Al!

 

[arawdy]

i have a similar question:
A man with mass M is sitting on a platform suspended by a system of cables and pulleys as shown . He is pulling on the cable with a force of magnitude F. The cables and pulleys are ideal (massless and frictionless), and the platform has negligible mass.

Find the magnitude of the minimum force that allows the man to move upward.

Base on what you have talked about, isn't it the F would be Mg/2 ??

or there are three ropes, so divide by 3?

 

[Doc Al]

Base on what you have talked about, isn't it the F would be Mg/2 ??

It depends on how many times the rope pulls up on the "man+platform". Realize that the pulley attached to the platform is part of the "man+platform". Note that the pulley arrangement in this problem is different than in the previous problem in this thread--here we have two pulleys connected. That makes a big difference.

or there are three ropes, so divide by 3?

Right!

 

[arawdy]

thank you, i get it!