Pulley Problem (with force upward)

In summary, the man exerts a force on the pulley shaft that makes the system go upward. The tension force to make the object not move is equal to the weight of the object. The normal force is zero to make the object not move and has no explanation at all.
  • #1
terryds
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Homework Statement


http://www.sumoware.com/images/temp/xzstefmnkmsctpxh.png
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

Homework Equations


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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  • #2
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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  • #3
NascentOxygen said:
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 ?
 
  • #4
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.
 
  • #5
NascentOxygen said:
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 ?
 
  • #6
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.
 
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  • #7
NascentOxygen said:
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 ?
 
  • #8
terryds said:
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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  • #9
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
 
  • #10
Chestermiller said:
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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  • #11
NascentOxygen said:
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 )
 
  • #12
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.
 
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  • #13
terryds said:
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
 
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1. What is a pulley and how does it work?

A pulley is a simple machine that consists of a wheel with a groove around its circumference and a rope or belt that runs through the groove. It is used to lift or move heavy objects by applying force in a different direction than the object's weight. When a force is applied to one end of the rope, the object on the other end is lifted or moved in the desired direction.

2. What is the purpose of using a pulley in a force upward problem?

In a force upward problem, the pulley serves as a mechanical advantage, allowing the force to be applied in a different direction than the object's weight. This makes it easier to lift or move the object, as the force required is reduced.

3. How is the force required to lift an object calculated in a pulley problem?

In a pulley problem with force upward, the force required to lift an object is equal to the weight of the object divided by the number of ropes supporting it. This is known as the mechanical advantage of the pulley system.

4. How do multiple pulleys affect the force required in a force upward problem?

Using multiple pulleys in a force upward problem can further reduce the required force by increasing the mechanical advantage of the system. Each additional pulley reduces the force required by a factor of the number of ropes supporting the object.

5. Are there any limitations to using a pulley in a force upward problem?

While pulleys can make lifting or moving objects easier, they also have limitations. The weight of the pulley itself, as well as friction in the system, can reduce the mechanical advantage and increase the force required. Additionally, pulleys can only change the direction of the force, not increase it beyond the applied force.

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