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Homework Help: Combined Mass

  1. Apr 25, 2013 #1
    Dear Forum Users,

    I am studying physics on my own, i am new to this forum, i am working every example in the book and i have a question. I hope that you are so kind to give me an explanation so that i can continue my chapter exercises.

    The question of the book is : What horizontal force must be applied to a large block of mass M shown in Figure P5.67 so that the tan blocks remain stationary relative to M ? Assume all surfaces and the pulley are frictionless. (Notice that the force exerted by the string accelerates M1)

    See attachement for picture.

    The calculation goes as follows.

    M1*g - T = 0 =>
    T = M2 * a =>
    M1*g = M2 * a => a = (M1*g)/(M2)
    F = TotalMass * a => in the book and on the internet the result is
    F = (M + M1 + M2) * (M1*g) / (M2)

    What i was wondering is if M1 is not connected by the string with M2, and the surface is frictionless, normally only block M would slide together with M1 and M2 would remain stationary and eventually fall off. The tension in the robe is provided by the gravitational force, no gravity
    no tension. So in the end result F only needs to accelerate M and M1 and not M2 because that
    is accelerated by gravity -> equation a = M1 * G / M2.

    So why is the result (M + M1 + M2) * (M1*g) / (M2) and not (M + M1) * (M1*g) / (M2).
    Because F is not responsible to give M2 that acceleration just for keeping M relative to M2 in rest.
    So they both need the same acceleration.

    Can you please help me with this reasoning.


    Attached Files:

  2. jcsd
  3. Apr 25, 2013 #2


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    Welcome to PF!

    Hi Jöran! Welcome to PF! :smile:
    I see what you mean.

    But no, for two reasons:

    i] suppose the three masses were fixed to each other …

    the result would be the same, wouldn't it? :wink:

    ii] this is an exercise in applying Newton's second law …

    you must always apply it to all the external forces on a particular body (or bodies)

    if you apply it to M and M1 combined, then you must include the external force (on M and M1) from the horizontal part of the rope …

    (M + M1)a = F - T = F - M2a,

    ie (M + M1 + M2)a = F :smile:
  4. Apr 25, 2013 #3


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    Hello Joran Verlaeck. Welcome to PF !

    If m1 and m2 are stationary relative M, then they all have the same acceleration, call it a .

    So the net force, F, on the system must be such that F = (m1 + m2 + M)a . It's simply the application of Newton's 2ND Law of Motion.
    Last edited: Apr 25, 2013
  5. Apr 25, 2013 #4


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    T = m1g = m2a. At the pulley, the string exerts a force T vertically and horizontally on M. The horizontal forces on M are therefore F, m1a from contact with m1 and m2a from the pulley.
  6. Apr 26, 2013 #5
    All of you guys,

    Thanks for presenting me the reasoning why this is the case.
    After i read the responses i released that i made a mistake by following my reasoning.
    I was thinking of the same physics situation only this time m1 rests on the floor, the outcome is like tiny-tim suggested the same. If so M1 feel the normal force and in reality there is really no tension in the robe, but once the object starts moving, and we want to make sure that M2 has no velocity relative to M, the tension in the robe gets m2.a and if a is exactly m1*g/m2 the normal force on m1 is zero. In this case you can see that F must be responsible to the motion of M2 and all the equations remains the same.

    Thanks for answering this topic.
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