# Pulley Problem (with force upward)

## Homework Statement

http://www.sumoware.com/images/temp/xzstefmnkmsctpxh.png [Broken]
A man pulls up a pulley shaft with force F (see the picture above)
Assume that the pulley and cord have no mass, and the mass of object 2 is bigger than the mass of object 1.
a) Determine the maximum normal force to make object 2 not moving.
b) Determine the tension force to make object 2 not moving

ΣF = ma[/B]

## The Attempt at a Solution

a) The normal force is a reaction force, right ?
So, What I see that the object 2 gives the weight force to the floor. So, the normal force is m.g , right ?
But, my book says that the normal force is zero to make the object 2 not moving and it has no explanation at all.

b) My book says that the tension force equals to the weight of the object 2.
But, I think that there must be a fictional force (pseudo force) exerts on object 2 and also object 1 (because the force the man exerts on the shaft pulley will make the system go upward, and the blocks must have an additional force downward).
So, I think that the tension will be T = W2 - F

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NascentOxygen
Staff Emeritus
Block 2 will leave the floor if the tension exceeds its weight. So the limiting tension T= m2.g

But, my book says that the normal force is zero to make the object 2 not moving and it has no explanation at all.
When tension up is about to lift body 2, there is zero force downward on the floor.

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Block 2 will leave the floor if the tension exceeds its weight. So the limiting tension T= m2.g

What about the force F ?
I think there must be a fictional force downward on the block 2, right ?
So, why isn't the tension the block 2 weight minus the force F ?

NascentOxygen
Staff Emeritus
Fictional?? What would that be?
Frictional? No.

Block 2 can't exert its full weight on the rope until the tension in the rope is about to lift block 2 off the floor.

Fictional?? What would that be?
Frictional? No.

Block 2 can't exert its full weight on the rope until the rope is about to lift block 2 off the floor.

Fictional force is a pseudo-force I mean
So, when a system is accelerated upward, the things in the system will accelerate downward (relative acceleration)
So, I think when the pulley shaft is accelerated upward, the block will have a downward gravitational acceleration and the downward relative acceleration.
Do I get this concept right ?

NascentOxygen
Staff Emeritus
I don't contemplate pseudoforces (except maybe when the bus brakes suddenly and I get thrown forwards). Too confusing for my simple mind.

Here, the pulley moving upwards accelerates the smaller block upwards. That seems straightforward.

terryds
I don't contemplate pseudoforces (except maybe when the bus brakes suddenly and I get thrown forwards). Too confusing for my simple mind.

Here, the pulley moving upwards accelerates the smaller block upwards. That seems straightforward.

Umm...
If the pulley shaft moves upwards, the smaller blocks will move downwards, right ? And, the smaller blocks will move upwards if the force F exerts on the blocks, right ? (tell me if I'm wrong or my logic is wrong)
∑F = m2 (g+arelative)
W - T = m2 (g+arelative)
T = m2 (g + arelative) / m2 g
T = m2 (g + ( m2 / ( m1 + m2 ) ) * F) / m2 g

What's wrong with this ?

NascentOxygen
Staff Emeritus
If the pulley shaft moves upwards, the smaller blocks will move downwards, right ?
Wrong. The rope length does not change, but I'm sure you know this.

T = m2g + m2a

Tmax = W

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terryds
Chestermiller
Mentor
Have you draw a free body diagram on (a) mass 1, (b) mass 2, and (c) the pulley?

If T is the tension in the wire and N is the upward normal force from the table, what are the force balance equations for (a) mass 1, (b) mass 2, and (c) the pulley?

Chet

Have you draw a free body diagram on (a) mass 1, (b) mass 2, and (c) the pulley?

If T is the tension in the wire and N is the upward normal force from the table, what are the force balance equations for (a) mass 1, (b) mass 2, and (c) the pulley?

Chet

I have some difficulties in drawing the diagram.
I'm doubt if the force F upward which is exerted by the man also have an impact on the blocks.
And, I think if the pulley shaft is exerted a force upward, then the blocks also have the upward acceleration due to the force F. So, I think the force F has an impact on the blocks. But, I'm still doubt on it. :L
Does force F also accelerate the two blocks ?
Is it T - W - F = ma ?

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Wrong. The rope length does not change, but I'm sure you know this.

T = m2g + m2a

Tmax = W

So, the force F upward has no impact on the block ?
So, what if the question is to determine the acceleration ?
Is it (a-g) ? ( a is the acceleration due to the force F or (m2/(m1+m2)) * F )

NascentOxygen
Staff Emeritus
One of the blocks remains on the floor, so it obviously doesn't accelerate.

Why don't you make up a simple pulley and rope system, just holding a drink bottle as the low-friction pulley and using a pair of weights joined with string, and try it out. You might discover a few things.

terryds
Chestermiller
Mentor
I have some difficulties in drawing the diagram.
I'm doubt if the force F upward which is exerted by the man also have an impact on the blocks.
And, I think if the pulley shaft is exerted a force upward, then the blocks also have the upward acceleration due to the force F. So, I think the force F has an impact on the blocks. But, I'm still doubt on it. :L
Does force F also accelerate the two blocks ?
Is it T - W - F = ma ?
Until you can draw these 3 free body diagrams, you are not going to be able to solve the problem. Ask yourself the question: Does the force F directly contact the two blocks? If the answer is "no", then it doesn't figure in the force balance equation on each of the blocks. But it does have an indirect impact on the blocks through the force balance on the pulley. You also need to understand that the acceleration of block 2 is zero, but the acceleration of block 1 is not zero. I'm going to give you a hint by writing the force balance equation for the pulley:

T + T = F

Now all you need to do is to write a force balance equation on each of the blocks.

Chet

terryds