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Homework Help: Pushing Blocks

  1. Apr 11, 2005 #1
    Four blocks of mass m = 10 kg are arranged as shown in the picture, on top of a frictionless table. A hand touching block 1 applies a force of F1h = 90 N to the right. The coefficient of friction between the blocks is sufficient to keep the blocks from moving with respect to each other.

    I have attached the picture to this problem:

    Oh , this problem is giving me a headache,

    Anybody have the slightest idea how to do this problem

    Attached Files:

  2. jcsd
  3. Apr 11, 2005 #2


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    You did not mention what was the problem. Was it the acceleration of the whole pile of blocks?

    As the three blocks above each other up can not move with respect to each other you can handle them as a single block of mass M=30 kg.

    Consider the forces acting separately on the first block with m=10 kg and on the second block (M=30 kg). The force of gravity cancels with the normal force from the table, you need to bother with the horizontal forces.
    There is the force of the hand acting on the first block and also the second block acts on it with a force of interaction f12.
    At the same time, only the force f21 from the block 1 acts on the second block. f21 = - f12 according to Newton's third law.

    Now you write down the equation of motion for both blocks, according to Newton's second law: mass times acceleration = the net force acting on the body. The two blocks touch each other, they will move with the same acceleration, a.

    ma = F1h + f12
    Ma = -f12.

    Add up the equations: the forces of interaction cancel and you get:

    a*(m+M)= F, that is a = F(m+M) = (90 N) /(40 kg) = 2.5 m/s^2 is the acceleration of the whole pile of blocks.

  4. Apr 11, 2005 #3
    The question is : What is the total force exerted on block 3 by block 2 ?
  5. Apr 12, 2005 #4


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    Well, we handle all blocks separately. There are horizontal forces and vertical ones. The vertical ones do not influence the acceleration, but they ensure the friction that makes all blocks move together.

    Vertical forces on the blocks: (upward direction is positive)

    block 1: gravity, G1= -m1*g ; normal force from the table, N1. The block does not move vertically, the net vertical force is zero: N1-m1*g=0.

    block 2: gravity G2=-m2g, normal force from the table N2, normal force from block 3: -N23. The block does not move vertically, so the net vertical force is zero N2-m2*g-N23=0

    block 3: gravity, G3=-m3*g. Normal force from block 2, (upward) N23, normal force from block 4 (downward) -N34. N23-N34-m3*g=0

    block 4: gravity G4=-m4*g, normal force from block 3 N34. N34-m4*g=0.

    Now you have to solve for all the normal forces:




    N2=m2*g + N23=(m2+m3+m4)*g.

    Now the horizontal forces, see the picture.

    Block 1 is pushed by Fp, block 2 is pushed by block 1, but according to Newton's third law, it pushes block 1 with a force of the same magnitude and
    opposite direction. Block 3 moves because of the friction between it and block 2. This is a positive force on it, but it acts on block 2 with a force of same magnnitude and opposite direction. Block 4 is accelerated by the frictional force between it and block 3, the same frictional force with opposite direction acts on block 3.

    So the forces in detail:

    block 1: push Fp, interaction from block 2 (negative): -f12. The block moves with acceleration a, so m1*a= Fp-f12.

    block 2: interaction from block 1 (positive) f12, friction between it and block 3 (negative) -fr23. m2*a=f12-fr23.

    block 3: friction from block 2 : f23, friction from block 4, - fr34.
    m3*a = fr23-fr34

    block 4: friction from block 3, fr34. m4*a = fr34.

    If you add up all equations, you get the acceleration:

    [tex] a = \frac{Fp}{m1+m2+m3+m4}=\frac{90}{40} = 2.5 m/s^2[/tex]

    plugging in the acceleration into the individual equations of motion, you get the forces of interaction.

    Now you can answer the question. the total force exerted on block 3 by block 2: It is the horizontal force -fr23=-20*2.5=-50 N, and the vertical force, N23=-(m2+m3)g=-20*9.8=-196 N. You can write it in vector form, or calculate the magnitude.

    Last edited: Jun 29, 2010
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