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How to solve this problem by using Newton's equations?

  1. Sep 8, 2015 #1
    upload_2015-9-8_19-6-5.png upload_2015-9-8_19-5-19.png They have solved this problem by energy method but i want to solve this by normal force equations but when i try to solve by this method suddenly a i had a doubt whether i have to solve the problem assuming that the point x2 always move in a straight line (this is easy to solve) or point x2 can move in a curve for solving this way , i think there should be more data available for this.
     
  2. jcsd
  3. Sep 8, 2015 #2
    No more data required.

    It does say when the system is in equilibrium so that means that nothing is moving.

    It also defines the length of string on either side to be equal, therefore x2 is always located on a straight line when the system is in equilibrium. However, you are right that when the system is not in equilibrium then x2 would move on a curve.
     
    Last edited: Sep 8, 2015
  4. Sep 8, 2015 #3
    so how to solve this one
     
  5. Sep 8, 2015 #4
    Didn't you say it was easy to solve?
     
  6. Sep 8, 2015 #5

    haruspex

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    Can you show the tension should be the same thoughout the string? Can you then show that the middle pulley must be half way between A and B?
     
  7. Sep 8, 2015 #6
    Yes i can solve , but i am not getting the same answer as them , so can you help me .
     
  8. Sep 8, 2015 #7
    Tension won't be same through out the string , it will be m1g between mass m1 and m2 and it will be m1g + m2g in between mass m2 and the point A .
    I think it depends on the m1, m2, x1 and x2 for certain values of this the middle pulley will always be half way between A and B. So that horizontal components acting on mass m2 cancel each other.
    Correct me if am wrong.
     
  9. Sep 9, 2015 #8
    Ah, maybe not so easy then. You'll need to show your work before anyone can help you.
     
  10. Sep 9, 2015 #9

    haruspex

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    if the tension is different in the string on one side of a pulley from what it is on the other, consider the torques acting on the pulley. What will happen?
     
  11. Sep 9, 2015 #10
    So far the point x2 to move in a straight line , there should be no horizontal component acting on it ,
    So at equilibrium force acting on the point x2 in the vertical direction is m1gcosθ1 + Tcosθ2= m2g where T= m1g+m2g , and θ1 is the angle between the line segments of point B and mass m1 and point B and mass m2 and θ2 is angle between the line segments point A and the mass m2 and vertical line drawn from A. here θ1 and θ2 should be equal since x2 moves vertically downwards , so answer i got for x1 is =(bm2- 2d(2m1 +m2 ))/m2
    but the answer they got is
    upload_2015-9-9_16-28-14.png
     
  12. Sep 9, 2015 #11
    rotate towards the higher force side
     
  13. Sep 9, 2015 #12

    haruspex

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    Right, but in this set up we are told it is in static equilibrium, so what does that tell you about the string tension?
     
  14. Sep 9, 2015 #13
    So far the point x2 to move in a straight line , there should be no horizontal component acting on it ,
    So at equilibrium force acting on the point x2 in the vertical direction is m1gcosθ1 + Tcosθ2= m2g where T= m1g+m2g , and θ1 is the angle between the line segments of point B and mass m1 and point B and mass m2 and θ2 is angle between the line segments point A and the mass m2 and vertical line drawn from A. here θ1 and θ2 should be equal since x2 moves vertically downwards , so answer i got for x1 is =(bm2- 2d(2m1 +m2 ))/m2
    but the answer they got is
    upload_2015-9-9_16-28-14-png.88428.png
    whether this is correct
     
  15. Sep 9, 2015 #14

    haruspex

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    I thought you were trying to use the force method, and this is in equilibrium, so nothing is moving. Why do you keep mentioning x2 moving in a straight line?
    But, yes, there cannot be any net horizontal force on the central pulley. The correct answer to my last post was "the tension is constant throughout the string". You denied that before but seem to be accepting it now, so let's move on.
    Since there is no net horizontal force at the central pulley, but there is a tension T each side, what does that tell you about the angles of the strings there?
    I don't understand how you get either of those equations. T most certainly is not equal to m1g+m2g.
     
  16. Sep 10, 2015 #15
    I
    I can't able to understand your statement can you please elaborate on that.
     
  17. Sep 10, 2015 #16

    haruspex

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    Which statement don't you understand?
    Do you agree now that the tension must be the same throughout the string?
    Consider the forces acting on the hanging pulley. There are three. What are their horizontal components? What equation does that give you?
     
  18. Sep 11, 2015 #17
    I don't understood how the tension in the string is constant throughout ? ,
    I will tell the idea behind how i got the value of my tension please tell whether this is correct ? , first i imagined the setup without the pulley M2 so the tension on the string is same and = m1g , now you are putting the pulley M2 on the string , so at this point of the pulley M2 will experience two forces acting on it one due to the mass m1(acting from pulley M2 to point B to restore to its original position) and due to mass m2 which is acting vertically downwards , so the tension on this point will be vector sum of these two forces , so depending on the points in the string there will be different tensions .
     
    Last edited: Sep 11, 2015
  19. Sep 11, 2015 #18

    haruspex

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    The flaw in your reasoning is that adding m2 won't increase the tension. It will merely redirect the tension. Instead of the tension being horizontal, it is now at an angle each side. That gives it a vertical component. Equilibrium is achieved when the vertical components of the tension each side balance the weight of m2. (If the weight is too great, equilibrium will not be achieved.)
    The moral is this: don't trust ad hoc arguments that seem logical. Go through the standard steps of drawing free body diagrams and writing out the force and torque balance equations.
     
    Last edited: Sep 11, 2015
  20. Sep 12, 2015 #19
    Thank you so much for your explanation , I got the same answer as them . Finally I am Happy.
     
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