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Force Problems Involving Two or More Objects

  1. Mar 11, 2007 #1
    1. Three blocks are in contact with each other on a frictionless horizontal surface.The masses of the blocks are m1 = 1 kg , m2 = 2 kg, m3 = 3 kg. A horizontal force F = 24 N is applied on m1.
    a. Find the net force on each block.
    b. Find magnitude of the contact forces between the blocks.

    2. Relevant equations
    F= ma

    3. The attempt at a solution
    I found the net acceleration, that equals to Fnet/(m1+m2+m3)
    and then for a) I just multiplied each mass by net acceleration to get the force. Like for mass 3, I multiplied net acceleration by mass 3

    and then for b)
    between the m1 and m2
    Net force on m1 is force applied - force exerted on it by m2
    and the force exerted by the forward masses equal to summ of m2+m1 divided by m1+m2+m3, and times total force<24 N>
    :frown: .....:cry: <I get lost after this>

    I am confused for both parts. Wondering if I am on right track?
  2. jcsd
  3. Mar 11, 2007 #2

    Doc Al

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    Staff: Mentor

    All good.

    To find the contact forces (there are only two: between m1 & m2 and between m2 & m3) consider the forces acting on each mass separately. You already know the net force, so: Analyze the forces on m1 to find the first contact force; analyze the forces on m2 to find the second contact force.

    Try it and see.
  4. Mar 11, 2007 #3
    so, I isolated the mass m1
    it is pushed background by the force from m2 that equals to net force of m2+net force of m3
    and it is pushed forward by the original net force that is applied at m1

    and I know both them
    and also the net force on m1

    so how can i find the contact force?
    Is it the force that pushes it backward?
  5. Mar 11, 2007 #4

    Doc Al

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    Staff: Mentor

    Keep it simple. What forces act on m1? Only two:
    (1) the applied force of 24 (to the right, say): call it +24
    (2) the contact force of m2 (to the left): call it -X

    So the net force on m1 (which you already calculated as m1a) must equal 24 - X. Solve for X!

    Now do the same for m2 to solve for the second contact force.

    (From Newton's 3rd law, you know that the force that m1 exerts on m2 must be equal and opposite the force that m2 exerts on m1.)
  6. Mar 11, 2007 #5
    X=24-net force on m1

    or X = net force on m2 + net force on m3

    they both are same?!
    that's why i am little confused... because I know X even before solving that equation.. and i get lost everytime about what I am trying to solve for
  7. Mar 12, 2007 #6

    Doc Al

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    Staff: Mentor

    Right. So now you know the contact force between m1 and m2.

    I don't know where this comes from.

    Instead, analyze the forces on m2:

    (1) Contact force from m1 (which we already know!): call it +X
    (2) Contact force from m3 (to the left): call it -Y

    So the net force on m2 (which you know is m2a) must be:
    X - Y = m2a

    Solve for Y and you're done.
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