Mechanics problem with two masses, a pulley and friction

In summary: It is good that we are no longer talking at cross purposes. Yes, M's own weight and the normal force from the table on which it rests are two things that act on M. There are others. Consider, for instance, block m2. Does it exert any force on M?But your primary concern is with tension and how it relates to M. Let us look more closely at that. Consider the pulley. What forces act on the pulley?The tension is the force of tension that is exerted on the rope by the pulley.
  • #1
Ahmed Ayman
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Can anyone please explain to me how can I calculate the "Tension" and "acceleration" of M in this question interms of M,m1,m2 and g?
I can't understand how M has an acceleration or why M is involved in the calculation of the tension
my solution was that acc of M is zero
and Tension = m1g x (m2/(m2+m1))
Thanks in advance :oldsmile:

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  • #2
Ahmed Ayman said:
Can anyone please explain to me how can I calculate the "Tension" and "acceleration" of M in this question interms of M,m1,m2 and g?
I can't understand how M has an acceleration or why M is involved in the calculation of the tension
my solution was that acc of M is zero
and Tension = m1g x (m2/(m2+m1))
Thanks in advance :oldsmile:

View attachment 233921
What is the tension in the rope (assuming M is not moving)? From a free body diagram of M (including the attached pulley), what are the forces acting on M?
 
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  • #3
Chestermiller said:
What is the tension in the rope (assuming M is not moving)? From a free body diagram of M (including the attached pulley), what are the forces acting on M?
the question didn't mention any external force, it just said that m1 can only move in the vertical direction plus that there is no friction in the whole system
 
  • #4
Ahmed Ayman said:
the question didn't mention any external force
The question does assume one external force in particular. But leave that to one side. What internal forces act on M?
 
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  • #5
jbriggs444 said:
The question does assume one external force in particular. But leave that to one side. What internal forces act on M?
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  • #6
Ahmed Ayman said:
[a graphic showing the problem]
The question was addressed to you, not to the textbook. You should be able to draw a set of free body diagrams, one for each object in the system. The question at hand is about the forces that act on M.
 
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  • #7
jbriggs444 said:
The question was addressed to you, not to the textbook.
that is the same question I'm trying to solve!
 
  • #8
Ahmed Ayman said:
that is the same question I'm trying to solve!
Yes, you have quoted the problem from the textbook. Yes, that is the problem that you are trying to solve. But you have been asked a question several times now about an inventory of forces that act on M. We are not asking for a numerical answer. We are just looking for you to identify those forces so that you could portray them in a free body diagram.

The goal is to be able to write equations that relate the various forces and accelerations that you identify. The first step is to identify the forces.
 
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  • #9
jbriggs444 said:
Yes, you have quoted the problem from the textbook. Yes, that is the problem that you are trying to solve. But you have been asked a question several times now about an inventory of forces that act on M. We are not asking for a numerical answer. We are just looking for you to identify those forces so that you could portray them in a free body diagram.

The goal is to be able to write equations that relate the various forces and accelerations that you identify. The first step is to identify the forces.
Oh sorry !
actually that is the main problem for me... I thought that M is affected by the weight and the normal force but I can't figure a relation between M and the tension.
sorry for not understanding your reply before!
 
  • #10
There are other forces. If M is accelerating to the right what about m1?

What about m2?
 
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  • #11
Ahmed Ayman said:
I thought that M is affected by the weight and the normal force but I can't figure a relation between M and the tension.
It is good that we are no longer talking at cross purposes. Yes, M's own weight and the normal force from the table on which it rests are two things that act on M. There are others. Consider, for instance, block m2. Does it exert any force on M?

But your primary concern is with tension and how it relates to M. Let us look more closely at that. Consider the pulley. What forces act on the pulley?
 
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  • #12
jbriggs444 said:
It is good that we are no longer talking at cross purposes. Yes, M's own weight and the normal force from the table on which it rests are two things that act on M. There are others. Consider, for instance, block m2. Does it exert any force on M?

But your primary concern is with tension and how it relates to M. Let us look more closely at that. Consider the pulley. What forces act on the pulley?
The tension ?
 
  • #13
CWatters said:
There are other forces. If M is accelerating to the right what about m1?

What about m2?
but how did you know that M is accelerating to the right?
 
  • #14
Ahmed Ayman said:
The tension ?
That's close. I am going to spoon feed you this part of the answer.

You have the tension from the rope segment leading off to the left. That results in a leftward force on the pulley. You have the tension from the rope segment leading down to m1. That results in a downward force on the pulley. You have the force from the attachment of the pulley to M.

Strictly speaking, "tension" is more a condition in a rope than it is an actual force. Tension results in a force at its point of attachment. That force is equal to the magnitude of the tension. It acts on the attachment point and has a direction pointed from the attachment point along the direction of the rope.

For an ideal pulley, the tension on the rope segment coming off on the one side is equal to the tension in the rope going in the other. However, those directions are almost always different, so pulleys are almost always under a net force due to the tension. You can treat a pulley as if the rope is "attached" to the pulley wheel at two points, one where it goes in and the other where it comes out.
 
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  • #15
jbriggs444 said:
That's close. I am going to spoon feed you this part of the answer.

You have the tension from the rope segment leading off to the left. That results in a leftward force on the pulley. You have the tension from the rope segment leading down to m1. That results in a downward force on the pulley. You have the force from the attachment of the pulley to M.

Strictly speaking, "tension" is more a condition in a rope than it is an actual force. Tension results in a force at its point of attachment. That force is equal to the magnitude of the tension. It acts on the attachment point and has a direction pointed from the attachment point along the direction of the rope.

For an ideal pulley, the tension on the rope segment coming off in the one side is equal to the tension in the rope going out the other. However, those directions are almost always different, so pulleys are almost always under a net force due to the tension. You can treat a pulley as if the rope is "attached" to the pulley wheel at two points, one where it goes in and the other where it comes out.
then the other force acting on M = the resultant of the two forces acting on the pulley ?
 
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  • #16
Ahmed Ayman said:
then the other force acting on M = the resultant of the two forces acting on the pulley ?
Yes.
 
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  • #17
jbriggs444 said:
Yes.
Ok thank you so much and sorry for wasting your time !
 
  • #18
The reason you are having so much trouble with this problem is that you have not drawn a free body diagram for the combination of M together with its attached pulley. Do you really feel that you have advanced to the point where you no longer need to use free body diagrams?
 
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  • #19
Chestermiller said:
The reason you are having so much trouble with this problem is that you have not drawn a free body diagram for the combination of M together with its attached pulley. Do you really feel that you have advanced to the point where you no longer need to use free body diagrams?
lol, I am really sorry, I thought I was able to imagine it but I will never forget to draw free body diagram again.. thank you for your help
 
  • #20
Ahmed Ayman said:
but how did you know that M is accelerating to the right?
There appears to be an outside force F acting on it to the right and no friction.
 
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  • #21
CWatters said:
There appears to be an outside force F acting on it to the right and no friction.
However, the full text in #5 reads "Let the force ##\vec{F}## be zero".
 
  • #22
Ahmed Ayman said:
lol, I am really sorry. I thought I was able to imagine it but I will never forget to draw free body diagram again.. thank you for your help
It's good that you recognize the error of your ways!

I'm betting that your instructor has advised you to draw a free-body diagram, your textbook says to draw a free-body diagram, and people in this thread suggested you draw a free-body diagram, yet you still neglected to follow this advice. Spend some time being metacognitive and ask yourself why you ignored all this advice and how you can avoid making similar mistakes in the future. It will help you become a better learner.
 
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  • #23
Ok,Thank you so much
 
  • #24
jbriggs444 said:
However, the full text in #5 reads "Let the force ##\vec{F}## be zero".
Sorry I missed that.
 

1. What is the purpose of including a pulley in a mechanics problem with two masses?

The pulley is used to change the direction of the force in the system. It allows for the masses to move in opposite directions while still being connected, which can make the problem easier to solve.

2. How does friction affect the motion of the masses in this mechanics problem?

Friction is a force that opposes motion, so it will act in the opposite direction of the motion of the masses. This means it will decrease the acceleration and speed of the masses, making the problem more complex to solve.

3. Can the masses in this mechanics problem ever reach a state of equilibrium?

Yes, it is possible for the masses to reach a state of equilibrium where the forces acting on them are balanced and there is no net acceleration. However, this depends on the initial conditions and the coefficients of friction in the system.

4. What equations or principles can be used to solve this type of mechanics problem?

The equations of motion (Newton's second law) and the principle of conservation of energy can be used to solve this type of problem. Additionally, the concept of torque and rotational motion may also be involved if the pulley has a mass and is rotating.

5. How does the inclusion of a pulley change the calculations and solutions in this mechanics problem?

The presence of a pulley will add additional variables and equations to the problem, making it more complex to solve. It may also require the use of rotational motion principles and the consideration of the pulley's mass and its effect on the system. However, it can also simplify the problem by changing the direction of forces and allowing for easier calculations.

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