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Pulleys and blocks

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    Case a) i)F=160N

    F is enough to overcome the static friction acting on block ##m_1##. Hence during motion kinetic friction acts and due to Newton's third law, an opposite and equal force acts on ##m_2##.

    For the motion of ##m_1## : ##F-\mu_k m_1g=m_1a \Rightarrow a=5.06 m/s^2## but this is wrong. :confused:
     

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  3. May 1, 2013 #2

    TSny

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    Are you sure? :smile:
     
  4. May 1, 2013 #3
    Static friction acting on ##m_1## is ##\mu_sm_1g=98N## which is less than 160N. I think I am missing something.
     
  5. May 1, 2013 #4

    TSny

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    Do you need 160 N of friction force to keep the block from slipping? Hint: The little block is going to accelerate even if it doesn't slip!
     
  6. May 1, 2013 #5
    Can you please distinguish "accelerate" and "slip"?
     
  7. May 1, 2013 #6

    haruspex

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    The upper block will accelerate relative to the ground even if it does not slip across (i.e. accelerate relative to) the lower block. Consider the forces on and acceleration of the upper block in an inertial frame.
     
  8. May 1, 2013 #7
    The forces acting in the direction of motion are F and the kinetic friction. I have already calculated the acceleration in the first post.
     
  9. May 1, 2013 #8

    ehild

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    First consider both blocks together as a system and see if static friction is enough to move the blocks together: It is given that the big block slides on the ground frictionless, while there is friction between the blocks, which tends to prevent relative motion. The system of the two blocks as whole is accelerated by the external force F. What is the common acceleration? Is static friction enough to provide the necessary force to both blocks?

    ehild
     
  10. May 2, 2013 #9
    I still don't see how the static friction is enough to prevent the relative motion between the blocks. Clearly, 160N is greater than 98N. :confused:

    If I assume that the static friction is enough, the common acceleration comes out to 3.2 m/s^2 which is correct as per the answer key.
     
  11. May 2, 2013 #10

    ehild

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    If the big block was fixed to the ground the force F would be enough to accelerate the small block on the surface of the big one. But the big block can move easily on the ground in lack of friction, and the small one initially rested on its surface. The static friction prevents relative motion - in any way. In this situation, that happens by moving the big block together with the small one. The external force accelerates the blocks together with a=3.2 m/s2. Check if the static friction is enough for both blocks to accelerate them with 3.2 m/s2.
    What happens when F is increased?

    ehild
     
  12. May 2, 2013 #11
    ##m_1a=64N## shows that static friction is enough to prevent the relative motion but why does the bigger block moves under the action F? I mean, if we make a free body diagram of ##m_2##, the two forces due to tension on pulley cancels out. :confused:
     
    Last edited: May 2, 2013
  13. May 2, 2013 #12

    ehild

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    m1a=64 N. It is the force needed to accelerate the small block with 3.2 m/s2. That force is the resultant of the tension and the force of friction, Fs, in opposite direction. What is the force of friction then?
    F accelerates the whole system, but the big block is accelerated by the static friction alone: 30a=Fs.
    Is the required static friction less than its maximum value?

    ehild
     
  14. May 2, 2013 #13

    haruspex

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    You've left out the frictional force from m1 on m2.
     
  15. May 2, 2013 #14
    ##F-f=64 \Rightarrow f=96N < 98N##, the static friction is enough to prevent slipping. Since the force which accelerates the bigger block is f, ##30a=96 \Rightarrow a=3.2 m/s^2##.

    But I am still confused, why can't we do it by considering the individual motion of each block? Why do we even need to consider the motion of the complete system and then the individual motion? :confused:
     
  16. May 2, 2013 #15

    haruspex

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    You can consider the whole system and one or other mass, or you can consider each mass. It should work either way.
     
  17. May 2, 2013 #16
    I tried to consider the motion of individual block but I ended up with a wrong answer.
     
  18. May 2, 2013 #17

    ehild

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    You write the equations for both individual parts of the system. There are external force(s) involved in the equations and also internal forces-forces of interaction between the parts. Such force is friction. But you can not know if it is static or kinetic.
    One is sure: the acceleration of the CM of the whole system is F(external)/(whole mass).

    If the friction is static, the parts move together with the same acceleration. That is one possible solution of the equations. You have to check if it is physically possible, that is the static friction can be big enough to accelerate both parts with the resulting acceleration.
    When the parts were in rest with respect to each other initially, always consider this case first.
    It it comes out that static friction is not enough to keep the parts together, change to the other method:

    The other possibility is that the parts move with respect to each other-from the beginning. The friction is kinematic then, and a well defined value. You can solve the equations, and get different accelerations for the parts. Again, you need to check if the solution is physically possible.

    Now try the solution with the other force.

    ehild
     
    Last edited: May 2, 2013
  19. May 2, 2013 #18
    How do I check if it is physically possible here? I make equations again for the individual motion, denoting the force of friction as f since I don't know its value. For the upper block, ##F-f=m_1a_1## and for the lower block, ##f=m_2a_2##. How do I relate ##a_1## and ##a_2##? If I assume they are equal, it gives the right answer but I am still confused as to why I need to consider they are equal. :confused:
     
  20. May 2, 2013 #19

    ehild

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    You can not relate a1 and a2 without knowing f. But you can assume they are equal. It is one possibility. From that, you get the required value of f.

    Do what I suggested, and try the other bigger force. Just to get the feeling how such problems can be treated.

    And you also can try some experiments with a toy truck and different loads...

    ehild
     
  21. May 2, 2013 #20
    When F=175N, the acceleration of the system is 175/50=3.5m/s^2.
    For the motion of ##m_1##, ##F-f=m_1a \Rightarrow f=105N## which is not possible as the maximum friction provided by the surface is 100 N (assuming g=10 m/s^2). So there will be relative motion between the blocks. Hence ##F-\mu_km_1g=m_1a_1 \Rightarrow a_1=5.75 m/s^2##.

    The only force acting on the bigger block is the frictional force. ##\mu_km_1g=m_2a_2 \Rightarrow a_2=2 m/s^2##.

    Does this look correct?

    Meanwhile, I was trying the Case b).
    As there is no net external horizontal force on the system, ##\displaystyle a_{cm,x}=0=\frac{m_1a_1+m_2a_2}{m_1+m_2} \Rightarrow a_2=\frac{-2a_1}{3}##.

    The direction to the right is positive. The forces acting on the upper block are the friction to the left and F to the right. ##F-f_k=m_1a_1##. Since F is enough to overcome the static friction, ##a_1=5 m/s^2## and therefore ##a_2=-10/3 m/s^2##. Does this look correct? Why was it safe to assume that F is sufficient to overcome the friction unlike the previous case?
     
    Last edited: May 2, 2013
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