Motion of a system that has two masses and three pulleys

  • #1

Homework Statement


IMG_20190314_004844.jpg

Homework Equations


F = ma

The Attempt at a Solution


I couldn't draw the freebody diagram. There's this weight of mass m, which is mg, downwards from m and there's tension T to upwards. This T affects mass M in the +x direction, but how could i find out the normal force? Also, the top left pulley is moving with mass M. Is there another tension T that impacts mass M in the +x direction?
Sorry for my bad English. Hope you could understand well enough to assist.
 

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Answers and Replies

  • #2
BvU
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HI,

Unfortunately, PF guidelines make it impossible for us to assist (note: to assist, not to answer) unless you post your own attempt...
 
  • #3
HI,

Unfortunately, PF guidelines make it impossible for us to assist (note: to assist, not to answer) unless you post your own attempt...

Hey, I edited. Could you please assist me? I kinda obsessed with that problem.
 
  • #4
BvU
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draw the freebody diagram
for m ? for M ? for one of the pulleys ?
another tension T that impacts mass M in the +x direction?
The one rope attached to m is not pulling in the horizontal direction (which I assume is your x direction ?)
 
  • #5
for m ? for M ? for one of the pulleys ?
The one rope attached to m is not pulling in the horizontal direction (which I assume is your x direction ?)
For both of the pulleys.
I actually mean the rope is pulling "M" in the direction of +x, which is right to the horizontal.
 
  • #6
BvU
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both of the pulleys
I see three of them :wink:, but I agree the one on the left is not very interesting.

For the others: A tension T to the left, the same tension T upward for the lower one and downward for the upper one. Since they are going nowhere, the sum of forces for each of the two must be zero -- meaning there is another force to keep them in place. Not so complicated, I would say :rolleyes:
 
  • #7
BvU
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I actually mean the rope is pulling "M" in the direction of +x
I thought I read an "m" there :confused:

It is. Twice.
 
  • #8
I see three of them :wink:, but I agree the one on the left is not very interesting.

For the others: A tension T to the left, the same tension T upward for the lower one and downward for the upper one. Since they are going nowhere, the sum of forces for each of the two must be zero -- meaning there is another force to keep them in place. Not so complicated, I would say :rolleyes:
Hey, I miswrote. I actually want to draw free body diagram for m and M. Not for the pulleys.
 
  • #9
BvU
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:-p

Well, if there is a tension T on the pulleys pulling them to the left, then on the lefthand side the rope is pulling with a tension T to the right !
 
  • #10
Poster has been warned to show their work in schoolwork threads
:-p

Well, if there is a tension T on the pulleys pulling them to the left, then on the lefthand side the rope is pulling with a tension T to the right !
I need to see the diagram, probably. With normal force and such. Because I can't see the simple explanation, unfortunately. :mad:
 
  • #11
BvU
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I need to see the diagram, probably
?:) PF is not about having your work done for you: you have to do it yourself. Shoot off your proposal and solicit comments !
 
  • #12
berkeman
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I need to see the diagram, probably. With normal force and such. Because I can't see the simple explanation, unfortunately. :mad:
We do not draw FBDs for you. Please show more effort on this schoolwork problem of yours, or the thread will be closed. Thank you.
 

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