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Forces acting on two blocks stacked one on another

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data
    I'll describe this without the numbers, values and such. The issue here is to understand the logic worked on this problem...well any problem is like that. Anyhow, this is the situation:

    A block of mass m1 is stacked on the surface of another block of mass m2 of width L, where m2>>m1. There is kinetic friction between those two blocks, static friction coefficient is not given. Block m2 is on a frictionless surface. An horizontal, constant force F is applied on block m1. The whole system of blocks moves.

    2. Relevant equations

    Sum of forces in x,y components;

    3. The attempt at a solution

    First thing: since there is a friction force, calculating the normal force in this case is important. For block m1,

    Fn_m1 - Fg_m1 + Fn_m2 = 0 -> Fn_m1 = Fg_m1 - F_m2;

    For block m2,

    Fn_m2 - Fg_m1 - Fg_m2 = 0 -> Fn_m2 = Fg_m1 + Fg_m2;

    The rest should just involve basic algebra, to obtain Fn_m1 and, therefore, the value of the kinetic friction force.

    Now, to work on the second part, the x-axis. Block m1:

    F - F_kin = m.a

    For block m2...here's where I think I don't know what I am doing. The picture of my Halliday book shows block m2 moving, so what kind of forces are acting on m2 in the x-axis? Is there a resulting force derived from the friction between the blocks? I've thought about this for a while and I'm just clueless. Not even sure if my logic for the forces on the y-axis is correct...
  2. jcsd
  3. Nov 5, 2007 #2


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    no, when you look at the forces acting on m1 inthe vertical direction, using a free body diagram, there are 2 forces acting: its weight down, and the normal force of m2 on m1, acting up. You've got an extra term in there.

    this is actually based on a FBD of both blocks together, and gives the correct value of Fn_m2

    yes, using the correct value for Fn_m1

    yes, correct

    Yes, draw a FBD of block 2. What force acts on m2 in the x direction (HINT: use your results from the FBD of block m1, and Newton 3).
  4. Nov 6, 2007 #3
    Okay, for m1 in y-axis, it'd be: Fn_m2 = Fg_m1, since the only normal force here is coming from the other block.
    As for m2, I don't really see why the forces drawn in a FBD for m2 for it don't match for my equation. There is the weight force from m1, and from m2 acting on it downwards and the normal force from the surface.

    For m2's forces in the x-axis, after some thinking, this is what I came up with: the friction force in m1 is a reaction against a force applied in m2. Therefore, for block m2,

    F_d = m.a, where F_d = F_kin in absolute values. Force F can't be drawn here, it was applied on block m1 only.
  5. Nov 6, 2007 #4


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    Okay there's a lot of subscripts here that lead to confusion, suffice to say that the normal force of m2 on m1 acts upward on m1 and is equal to the weight of blocK m1.
    when you draw a FBD of m2, there are 3 forces acting on m2 in the vertical direction: The weight force of m2, the normal force upward from the surface, and the normal force downward from m1 on m2. The weight of m1 does not enter into the FBD of m2, although it happens in this problem to be the same as the normal force acting downward from m1 on m2.

    yes, correct, but be sure in your FBD of m2 that your equation reads F_kin = m2a2, and note that a2 is not the same as a1, I'm not sure if this is clear to you, is it?
  6. Nov 6, 2007 #5
    Alright got it. As for the last line, I can understand that a2 << a1, because, in this case m1<<m2, and F (force applied on m1) >> F_kin. A force of small magnitude on a block of great mass accelerates the block at a much slower rate than a block of small mass pushed by a force of great magnitude.
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