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Masses accelerating

  1. Aug 8, 2015 #1
    1. The problem statement, all variables and given/known data
    All surfaces are smooth. mass m1 is attached to m3 with a smooth rail, it can't detach.
    What are the accelerations of the masses relative to the floor.

    2. Relevant equations
    Conservation of momentum: ##m_1v_1+m_2v_2=0##

    3. The attempt at a solution
    a3 and a2are the acceleration of masses m3 and m2 relative to the floor. ##a_2'## is m2's acceleration relative to m3.
    In the accelerating frame of m3:
    $$\left\{ \begin{array} {l} m_2a_3+T=m_2a'_2 \\ m_1g-T=m_1a'_2 \end{array} \right.$$
    Accelerations relative to the floor: ##a_2=-a_3+a'_2##
    These equations yield a relation between a3 and a2 that includes g.
    If i consider relative to the floor, from conservation of momentum:
    $$(m_1+m_3)v_3+m_2v_2=0\Rightarrow (m_1+m_3)a_3+m_2a_2=0$$
    From this we get relation without g.
    Snap1.jpg
     
  2. jcsd
  3. Aug 8, 2015 #2

    Titan97

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    Are you sure you can conserve momentum of the system? The centre of mass of the system has an acceleration.
     
  4. Aug 8, 2015 #3
    Post edited .
     
  5. Aug 8, 2015 #4
    Yes , this is correct ( as in , your comment ) .
    Also , another hint : Is there any force acting on m3 ? So what is it's acceleration ?
    Does this show how you have gone wrong again ?
     
  6. Aug 8, 2015 #5
    I don't think that it matters that the masses accelerate, there is still conservation of momentum since there aren't any external forces in the x direction, so the whole of the masses can't move. the C.O.M remains in place, and from that i also got the same answer.
    The only force i see that acts on m3 is the tension T. relative to the floor:
    $$\left\{ \begin{array}{l} m_1g-T=m_1a_2 \\ T=m_2a_2 \end{array}\right.\Rightarrow a_2=\frac{m_1}{m_1+m_2}g$$
    The tension T acts on m2 and on m3:
    $$F=ma\rightarrow m_2a_2=(m_1+m_3)a_3$$
    This is the same as momentum conservation, same as before.
     
  7. Aug 8, 2015 #6
    No , you can't conserve momentum . Firstly , there is a net force in the x - direction , from the tension acting on m2 .

    No force acts on m3 . It remains stationary .
     
  8. Aug 8, 2015 #7

    haruspex

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    Wrong on both counts.
    No masses are given for the wheels, so we should take them as massless. The consequence is that it will be just as though m3 is sliding on the ground without friction.
    As Karol posted, there are no external horizontal forces on the m1+m2+m3 system, so its mass centre stays put, and conservation of momentum can be used. However, the trap is that m3 will lose contact with m1.

    Karol, I'm not sure what question you are asking with your post. You ended up with two equations and two unknowns, which is fine if the equations are correct. What stopped you at that point?
     
  9. Aug 8, 2015 #8
    Why would m3 start moving ? There is no force acting on it .
    The tension acting on m2 ?
     
  10. Aug 8, 2015 #9

    haruspex

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    The string exerts a normal force around the curve of the pulley. The net of this will be a force acting down and left at 45 degrees.
    I don't understand your question. are you suggesting that this is a force external to the m1+m2+m3 system?
     
  11. Aug 8, 2015 #10
    I am very sorry i didn't draw it more clearly and didn't explain the conditions.
    Mass m1 can't detach from m3 because it slides on a rail that is attached to m3!
     
  12. Aug 8, 2015 #11

    haruspex

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    No, my mistake - you did indeed write that in the OP.
    So, what's your issue?
     
  13. Aug 9, 2015 #12
    The issue is that when i solve in the non inertial frame attached to m3:
    $$\left\{ \begin{array} {l} m_2a_3+T=m_2a'_2 \\ m_1g-T=m_1a'_2 \end{array} \right.$$
    Accelerations relative to the floor: ##a_2=-a_3+a'_2\rightarrow a'_2=a_2+a_3##
    With conservation of momentum: ##(m_1+m_3)a_3+m_2a_2=0## i get:
    $$a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)-m_1m_2}$$
    Relative to the floor:
    $$\left\{ \begin{array}{l} m_1g-T=m_1a_2 \\ T=m_2a_2 \end{array}\right.\Rightarrow a_2=\frac{m_1}{m_1+m_2}g$$
    With conservation of momentum: ##(m_1+m_3)a_3+m_2a_2=0## i get:
    $$a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)}$$
    The denominator is different
     
  14. Aug 9, 2015 #13

    haruspex

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    That does not look right to me.
     
  15. Aug 10, 2015 #14
    Relative to the floor:
    $$\left\{ \begin{array}{l} m_1g-T=m_1a'_2 \\ T=m_2a_2 \\ a_2=a_3+a'_2 \end{array}\right.\Rightarrow a_2=\frac{m_1g-m_1a_3}{m_1+m_2}g$$
    Conservation of momentum:
    $$(m_1+m_3)a_3+m_2a_2=0$$
    $$\Rightarrow a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)-m_1m_2}$$
     
    Last edited: Aug 10, 2015
  16. Aug 10, 2015 #15

    haruspex

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    All good now?
     
  17. Aug 10, 2015 #16
    Thank you very much Haruspex and the others
     
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