Two pulleys lifting a mass in parallel

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Hamiltonian
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Homework Statement
In the arrangement shown in the figure, the ends P and Q of an un-stretchable string move downwards with uniform speed U. Pulleys A and B are fixed. The mass M moves upwards with a speed
Relevant Equations
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2pulley.png
if the ends P and Q are being pulled down with a uniform speed its acceleration is zero and hence the Tension in the string will also be zero and if this is the case which force will make the block of mass M rise? is this a fatal flaw in the question?
 
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Hamiltonian299792458 said:
Homework Statement:: In the arrangement shown in the figure, the ends P and Q of an un-stretchable string move downwards with uniform speed U. Pulleys A and B are fixed. The mass M moves upwards with a speed
Relevant Equations:: -

View attachment 267684 if the ends P and Q are being pulled down with a uniform speed its acceleration is zero and hence the Tension in the string will also be zero and if this is the case which force will make the block of mass M rise? is this a fatal flaw in the question?
If the pulleys were so close together that ##\theta = 0## then you would be right about there being no acceleration of the mass vertically (constant speed). But since they show it being non-zero, what happens to the motion of the mass upward as it gets closer to the pulleys? How does ##\theta## change?

EDIT -- updated my reply above in bold to clarify my point. Thank you to @hutchphd for pointing out my ambiguity via a PM.
 
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berkeman said:
If the pulleys were so close together that ##\theta = 0## then you would be right. But since they show it being non-zero, what happens to the motion of the mass upward as it gets closer to the pulleys? How does ##\theta## change?
##\theta## increases with time, and since the velocity of the block depends on ##\theta## it accelerates upwards? so due to this, a tension is created in the string. so if a person was to pull on the points P and Q he would feel a tension even though he is pulling at a constant speed?
 
Hamiltonian299792458 said:
##\theta## increases with time, and since the velocity of the block depends on ##\theta## it accelerates upwards? so due to this, a tension is created in the string. so if a person was to pull on the points P and Q he would feel a tension even though he is pulling at a constant speed?
Since pulling the string downwards lifts the mass, there will have to be a tension in the string. That tension will depend on the mass and the angles.

As the mass gets higher up, does it speed up or slow down?
 
berkeman said:
Since pulling the string downwards lifts the mass, there will have to be a tension in the string. That tension will depend on the mass and the angles.

As the mass gets higher up, does it speed up or slow down?
intuitively I feel it should slow down( i may be wrong). I am unable to come up with an equation involving U and ##\theta## hence I cannot say for sure
 
FactChecker said:
Are you ignoring gravity?
no, I am not ignoring gravity. i am talking about the speed at which the ends P and Q are pulled at
 
Hamiltonian299792458 said:
no, I am not ignoring gravity. i am talking about the speed at which the ends P and Q are pulled at
Then doesn't there need to be tension in the string just to keep the mass from falling? Is that not supposed to be addressed in this problem?
PS. Your "Homework Statement" does not say what you are supposed to determine.
 
FactChecker said:
Then doesn't there need to be tension in the string just to keep the mass from falling? Is that not supposed to be addressed in this problem?
PS. Your "Homework Statement" does not say what you are supposed to determine.
we need to find the speed of the block M in terms of theta and U. the tension is inst mentioned in this problem but I wanted to know if a situation like this is actually possible or not?
 
Hamiltonian299792458 said:
we need to find the speed of the block M in terms of theta and U. the tension is inst mentioned in this problem but I wanted to know if a situation like this is actually possible or not?
Yes, it is possible. To be more accurate, you should say that there is no additional tension due to acceleration. There is still the tension needed to oppose the force of gravity, which would otherwise slow the rise of the mass, M.
 
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FactChecker said:
Yes, it is possible. To be more accurate, you should say that there is no additional tension due to acceleration. There is still the tension needed to oppose the force of gravity, which would otherwise slow the rise of the mass, M.
There most certainly would be additional tension due to the upward acceleration of the mass. In order to maintain a uniform speed on the ropes, the mass M must accelerate upward. In addition to gaining height (and increasing its potential energy), it will be gaining speed (and increasing its kinetic energy). Both of those require that the energy come from somewhere. It is coming from the work done by the ropes. The tension on the ropes needs to be enough to power not only the increasing potential energy but the increasing kinetic energy as well.
 
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Hamiltonian299792458 said:
Homework Statement:: In the arrangement shown in the figure, the ends P and Q of an un-stretchable string move downwards with uniform speed U. Pulleys A and B are fixed. The mass M moves upwards with a speed
Is that the whole question?
If so, I think all they are asking for is the instantaneous speed at the illustrated position. This is purely a kinematic problem.
But maybe I misunderstand your difficulty.
 
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There may be no question about the homework problem as stated. This side-question of tension on the rope is answered by considering the force of gravity and the acceleration of the mass due to the changing geometry of the pulley system.
 
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