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Find the force required to prevent the mass m2 from descending

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    For the system shown below, find the force required to prevent the mass m2 from descending. Use the values of m1 = 9.00kg, m2 = 1.75kg and m3 = 1.000 kg. Assume that all the surfaces are frictionless and the ropes do not stretch (also, in the figure, assume that mass m2 is flush against m1).

    2. Relevant equations
    F = ma
    W = mg

    3. The attempt at a solution
    I've drawn the free body diagrams for each of the masses.

    For mass m2, i've drawn the tension (upwards) and the weight (downwards).

    For mass m3, i've drawn the force that m1 exerts on m3, the weight (but these two wouldn't count as the mass isn't moving vertically), and the tension directed to the pulley (or must it be in the opposite direction?).

    For mass m3, i've drawn the force that the surface exerts on m1, the force that exerts m3 on m1, the force that we are applying and it's weight (also, i didn't take into accoutn the vertical forces for m1 and m3).

    By this, i concluded that m3 is also moving with the acceleration that the force on m1 provides, but the tension must cancel out the acceleration (force) so that m2 doesn't move.

    How can i proceed from this information or have i made a mistake in my assumptions?

    Attached Files:

    Last edited: Oct 2, 2013
  2. jcsd
  3. Oct 2, 2013 #2
    You need to apply a force to M3 that will balance the force of gravity on M1.
    However, F will be applied to all three masses. So F will be proportionally stronger than the force needed on M3 (M1+M2+M3)/M3.

    I think that should get you started.
  4. Oct 6, 2013 #3
    So pretty much i have the free body diagrams for the masses.

    In the case of M2, we would only have the tension and the weight of the block.

    In the case of M1, we would have the force (the one we have to exert), the weight of the block, the force that M3 exerts on M1 and the normal force.

    Finally, for M3, we would have the force that M1 exerts on M3, the tension of the string and the weight of the block.

    Where does that other force on M3 come from? I was thinking of writing another force on the FBD of M1 (which is the force that the pulley exerts on M1).
  5. Oct 6, 2013 #4
    Since M2 is flushed against M1,both exert a normal contact force(N) on each other .You need to take that force into account .

    This N opposes the horizontal motion of M1 ,but accelerates M2 horizontally.

    Pulley doesnt exert force on M1.It is part of M1.You need to take into account forces acting on the pulley as if they are acting on M1 .

    What is that which exerts force on the pulley ?
  6. Oct 6, 2013 #5
    What does being "flush" mean? I'm sorry, my english is not that great.
  7. Oct 6, 2013 #6


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    Staff: Mentor

    "flush" in this case means they are touching (so m2 won't swing on its rope to strike m1 when F starts to pull on the system).
  8. Oct 6, 2013 #7
    Okay, so from what i understand i think that the tension on the rope is the weight of the mass 2 (as it doesn't move up or down). Thus, it would also mean that the tension that the mass 3 experiences is the same. Up to that point i have the following FBDs:

    For the mass 3, we have the force that 1 exerts on 3, the tension of the rope, and its weight. For the mass 2, we have the tension, the force that 1 exerts on 2 and its weight. Finally, for the mass 1, we have the normal force, its weight, the force that 3 exerts on 1, the force that 2 exerts on 1 and the force that pulls the system.

    How can i proceed with that information? I'm not sure on how to proceed.
  9. Oct 6, 2013 #8


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    Good. So, what sort of motion of m3 would result in it producing that tension in the rope?
  10. Oct 7, 2013 #9
    you are missing something. the thing is pulley is part of mass 1, that means the force which acts on the pulley acts on mass. And, yes I can assure you the some forces are acting on the pulley. I have done this problem in the past, and I even remember the answer.
  11. Oct 7, 2013 #10

    So we would have two additional forces in the FBD that correspond to the tension? Then what would be the net force on the mass 1? (i've thinking that the accelerations should be equal).


    I'm not sure if i'm catching on with your question. Isn't in this case the motion in the forward direction? (however are the pulleys stationary or do they travel at the same acceleration as does the mass 1?).
  12. Oct 7, 2013 #11


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    Staff: Mentor

    All of the masses remain stationary with respect to each other, and the pulley does not turn. But the system as a whole is being accelerated (with respect to the motionless table frame) by the applied external force.
  13. Oct 7, 2013 #12
    If all of the masses travel at the same acceleration, then wouldn't the mass 3 movement in the forward direction be provided by the tension of the rope? And thus the one in the mass 2 provided by the contact force of mass 1 and mass 2? Or am i missing any other force?

    At least i've got clear that the forces in the y direction cancel out as there isn't any movement in that direction.
  14. Oct 7, 2013 #13


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    Staff: Mentor

    That's a reasonable summary of the situation. So what acceleration must mass m3 experience in order to match the tension in the rope?

    A big hint: To an observer moving along with an accelerated frame of reference, the observer "feels" a force acting that is indistinguishable from a "gravitational" acceleration acting on him. Effectively, he experiences a weight due to the accelerated motion of the whole system. That force is directed directly opposite the direction of acceleration of the whole system. It is a result of inertial forces.

    In this case m3 is the "observer". What acceleration will make his "weight" to the left equal to the tension force in the rope?
  15. Oct 7, 2013 #14
    So then there is sort of a "inertial force" that we have to take into account with accelerating frames. The acceleration would then be equal to the one experienced in m1? (i'm not so sure about this one).
  16. Oct 7, 2013 #15


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    That's the basic premise.

    All masses resist a change in motion, hence Newtons 1st and 2nd laws. Newton's 3rd law tells you that when something pushes on a mass, the mass pushes back. All together, this is the expression of inertia.

    In an accelerating frame of reference the inertia of a mass expresses itself as a "weight" in that frame. It's just as if there were a gravitational force in the form of a uniform gravitational field acting. If the frame is accelerating with acceleration "a", then "a" is the acceleration due to "gravity" that the masses experience in that frame.

    In this problem we are saying that all three masses are remaining motionless with respect to each other (in a frame of reference co-moving with them), and the whole set is being accelerated by some external force F causing them all to accelerate to the right with some acceleration a. This acceleration results in a "weight" for each mass that is directed to the left. This is important for masses m2 and m3. In the moving frame, while m2 is being pulled downwards by "real" gravity according to f = mg, mass m3 is being pulled leftwards by pseudo-gravity f = ma. The body of mass m1 prevents m3 from responding to g, and m2 from responding to a (m3 is resting on top of m1 and can't accelerate downwards, while block m2 is resting flush against the side of m1 and can't accelerate leftwards).

    So the question to ask yourself is, what value of a will make the pseudo-gravitational force on m3 balance the real gravitational force acting on m2 being transmitted via the rope and pulley? Then work out the force F that will cause that acceleration for the whole system.
  17. Oct 7, 2013 #16
    So i finally got the answer!

    What i did was the following: i assumed that all of the three masses were moving at the same acceleration (then, their net forces would be given by the mass multiplied by the acceleration). For mass 3, we have that the tension gives this net force. On the other hand, for mass 2 the contact force of 1 on 2 gives the net force. For the mass 1, we have the tension, the force and the contact force of 2 on 1.

    Then, we knew that the net force acting on mass 3 is the tension acting on it, so m3a = m2g (remember the value of the tension is the same). Then i worked out that a = (m2g)/(m3).

    Now, for the other mass 1, i assumed that there were the following forces acting on the body:

    The contact force of mass 2 with mass 1 (which is m2a), the tension of the rope (which in this case would be in the other direction) and the force. This would equal a net force, which is given by m1a. Then i cleared the equations and figured out that the force was given by:

    F = (gm2(m1+m2+m3)/(m3), which with plugging in the numbers, we find that the result is 202N.

    Thank you gneill!
  18. Oct 7, 2013 #17


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    Staff: Mentor

    Nice. :smile:
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