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Condition satisified when the body does not slide

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data

    A sledge of mass m1 is pulled horizontally with a force F. On the sledge there is a body of mass m2 that can slide on the horizontal platform of the sledge with the friction coefficient μ. Another sledge of mass m3 is tied with a horizontal string of the body m2. Between the sledges and the snow the friction is negligible. Find the condition that is satisfied when the
    m2 body does not slide.

    2. Relevant equations
    fs = μN

    T = [m2/(m1 + m2)]*F

    3. The attempt at a solution
    I assumed that μ is the coefficient of static friction. If the m2 body is not to slide, the static friction between this body and the m1 sledge must be bigger than the tension in the string.
    The tension in the string should be:

    T = [m3/(m1 + m2 + m3)] * F

    and the static friction:

    f = m2

    m2gμ > [m3/(m1 + m2 + m3)] * F

    This gives:

    μ > F/m2g * [m3/(m1 + m2 + m3)]

    but the book answers say that:

    μ > F/m2g * [(m3 + m2)/(m1 + m2 + m3)]

    What is wrong?
  2. jcsd
  3. Mar 19, 2017 #2


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    Draw a free body diagram for m2 and set up the second law for m2. This will give you the correct relation between the static friction force, the tension, and the acceleration.
  4. Mar 20, 2017 #3
    The acceleration of the system is F/(m1 + m2 + m3). This means that horizontally, in the positive direction, on m2 acts the force:

    F' = m2 F/(m1 + m2 + m3)

    The tension acts horizontally in the negative direction:

    T = - m3F/(m1 + m2 + m3)

    So, on m2 acts horizontally this force:

    Fr = F(m2 - m3)/(m1 + m2 + m3)

    which can be positive or negative. The friction should compensate for the resultant force, so it should be positive and equal in magnitude with the tension.

    But, the solution says that
    μ > F/m2g * [(m2 + m3)(m1 + m2 + m3)]

    which in my opinion means that the friction should be bigger that the F' + T. I don't understand where this sum comes.
    Last edited: Mar 20, 2017
  5. Mar 20, 2017 #4


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    Yes, that represents m2a. So it must equal the net force acting horizontally on m2. That is, it equals ∑Fx.


    There is a sign error here. Start with ∑Fx = m2ax. On the left side, add all the x-components of the forces acting on m2, taking into account the signs of the x-components. Then solve for Fr.

    If the friction force is equal in magnitude to the tension force (but opposite in direction), would m2 have any acceleration?
  6. Mar 20, 2017 #5
    Oh. Now I get it. The force acting on m2 is the resultant of tension and friction, the friction being in the positive direction:

    -m3F/(m1 + m2 + m3) + m2gμ = m2F/(m1 + m2 + m3)

    Thank you.
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