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Hanging mass force

  1. Feb 18, 2008 #1
    A box (m = 20 kg) is sitting on a horizontal surface. It is connected to a massless hook
    by a light string passing over a massless pulley wheel. The coefficients of friction
    between the box and the surface are 0.50 (static) and 0.30 (kinetic). On top of m is a
    second box M of mass 20 kg. The coefficients of friction between the boxes m and M
    are 0.80 (static) and 0.60 (kinetic). How much weight needs to be added to the hook
    until the box just begins to move.
    A) 10 kg B) 98 kg C) 20 kg D) 16 kg E) 200 kg

    So the total mass of m= 40kg
    so fs=40*9.80*.5
    so fs= 196N
    thus the force pulling down on the second mass has to equal fs
    the force pulling down is mg
    so mg=196N

    I do realize that on the second mass hanging over the table has both a tension force and a mg force, do you have to take the tension into account? I wouldnt think so because we are trying to figure out how much mass is need to start the first mass moving, so the tension doesnt matter, its just the force that carries the mg force from the second mass to the first mass. Making mg=fs

    Isnt the force of the system = 0 before it starts moving thus 0= mg -fs or mg=fs???
    I do not like my way of solving this problem. How would you guys solve it??

    Last edited: Feb 19, 2008
  2. jcsd
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