Pulley with string attached to ground

[khkwang]

Not exactly a homework question, but I need to know this before I can answer the question

Homework Statement

There's a pulley suspended from a rope A. Hanging from the pulley from one side is a mass m. "Hanging" off the other side of a pulley is simply the same rope (which is attached to the mass) (rope B), attached to the ground.

Is the tension on rope A simply mg?

Homework Equations

none

The Attempt at a Solution

I know the tension on rope B would be mg. But I'm wondering about rope A. I don't know whether to count the side of rope B attached to the ground as a contributing factor in the tension of rope A.

The way I see it is that since rope B is attached to the ground, the only thing that moves if one were to pull on rope A would be mass m. Thus the only work done is on mass m, and the only force required to move mass m would be needed: mg.

So then the tension on both ropes would be mg?

 

[ehild]

The work done on mass m if rope B is pulled does not concern the rope A. The centre of the pulley where it is suspended remains stationary. The resultant of the forces acting at the pulley must cancel. The tension is the same along a rope. Rope B pulls the pulley at both sides.

ehild

 

[khkwang]

I'm confused about your reply and I don't know if the confusion is on my side or yours... just to clarify by question, I attached a diagram.

I'm wondering if rope A were pulled not rope B

 

[ehild]

If you pull rope A you have to pull both the mass and the Earth. Rope B is fixed to Earth.
If you prefer thinking in terms of work, find out how much the mass will rise if you pull up the pulley by a certain length Δy. Turn the figure upside down. Is not it familiar from your textbook?

If you cut rope B and fix the free end to a peg on the pulley the mass will move to a stationary position hanging bellow the centre of the pulley. In that case, the tension would be mg in the rope A.

ehild

 

[khkwang]

I'm feeling pretty dense right now... I don't get exactly what you're trying to imply (I do you know you want me to come to my own conclusion though, rather than spell it out for me).

I'm taking it as the peg attached to the end of the rope is equivalent to the rope attached to the ground in the sense that in both cases the Δy of the mass is equivalent to the Δy of the pulley. So in both cases the tension on rope A is mg.

 

[ehild]

I'm taking it as the peg attached to the end of the rope is equivalent to the rope attached to the ground in the sense that in both cases the Δy of the mass is equivalent to the Δy of the pulley. So in both cases the tension on rope A is mg.

No, the displacement of the mass is not the same as the displacement of the pulley when the rope B is fixed to the ground.

See this place. http://simple.wikipedia.org/wiki/Pulley

Your system is similar to an upside down movable pulley.

I must not tell you the solution. I only can help you to find it.

ehild

 

[khkwang]

Oh no, of course you can't tell me the answer; that would defeat the purpose. I hope you didn't take the statement in my last reply as resentment!

Okay, I'll try again.

Since the mass is at rest, the tension all along rope B is equal. So the tension to the left of the pulley and to the right of the pulley are the same. Since the tension on the right is mg, then the one on the left must also be mg.

Since rope A is holding both tensions up, it must then have a tension equal to the sum of the tensions acting against it. Therefore T_A = 2mg.

Is that right?

 

[ehild]

It is right now.

ehild

 

[khkwang]

Thanks for your help and patience.

Cheers