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What force is Block A exerting on Block B?

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data

    This is a question that the teacher gives every week that we do in class. He gives the parameters without the actual numbers a week before so we can prepare.

    The teacher will give:

    - the masses for both blocks (different)
    - the coefficient of friction for both blocks (different)
    - the applied force on Block A

    Question (what we have to figure out): What force is Block A exerting on Block B?

    http://img580.imageshack.us/img580/7968/85935850.png" [Broken]

    2. Relevant equations

    Symbols weren't working for me but I'm assuming these equations are relevant:

    coefficient of friction = Net Force / Normal Force


    Fnet = mass * acceleration

    3. The attempt at a solution

    I'm not sure how to go about at this. I would have to specify some numbers before I try. I would have to find all the forces acting on both objects first (one force would be the mass, one force would be the friction, and one would be the applied force for Block B). After that, I have no idea what to do.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 22, 2011 #2


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    When you look at both objects sliding together as a system, you have, in the x direction, the applied forces and the friction forces. Mass is not a force. Knowing these forces and the mass, you can calculate the acceleration of the blocks as they move together with the same acceleration.
    To calculate the force between the blocks, you have to draw a free body diagram of one of the blocks and identify all forces acting on it, then apply Newton's law again to that block to solve for the unknown force. You should look at the other block also in the same manner, as a check on your work.
  4. Oct 23, 2011 #3
    Alright, so I tried specifying numbers on my own. Here's my go at it:

    Block A mass = 8 kg
    Block B mass = 20 kg
    Coefficient of kinetic friction for Block A = 0.20
    Coefficient of kinetic friction for Block B = 0.40
    Applied Force = 200 N

    Find Normal Force
    Block A
    F = ma
    F = 8kg x 9.8
    F = 78.4 N
    F = (78.4N)(0.20) - taking coefficient of friction into consideration
    F = 15.68 N
    Block B
    F = ma
    F = 20kg x 9.8
    F = 196 N
    F = (196N)(0.40)
    F = 78.4 N
    Find Acceleration
    Total normal force = 78.4 N + 15.68 N = 94.1 N
    Convert to kg = 94.1 N/9.8 = 9.60 kg
    a = F/m
    a = 200N/9.6kg
    a = 20.83m/s^2 [right, assuming applied force is to the right]
    Force of Block A on Block B
    F = ma
    F = (8kg)(20.83m/s^2)
    F = 166.64 N [E]

    I know I completely messed up one way or another. Would this be the right way, if not, can someone please correct me?
  5. Oct 24, 2011 #4
    I probably didn't do this right.
  6. Oct 24, 2011 #5
    try to draw a free body diagram of both the masses seperately. assume the accln. to be the same for both the masses, because otherwise they would break contact (i.e. if B has more acln. than that of A, also think why it is not possible that B has less accln. than that of A) and hence there will be no force btwn. them.

    the forces on A:

    A1. applied force F
    A2. Friction opposing the motion.
    A3. force exerted by B on A, also opposing motion.

    add the forces vectorially and equate the result with mass X accln. of A.

    the forces on B:
    B1. Friction opposing the motion.
    B2. force exerted by A on B, this time favouring the motion.
    add the forces vectorially and equate the result with mass X accln. of B.

    now from Newton's 3rd law put A3 = B2

    now u have 3 eqn.s with three unknowns, assuming the accln. of the masses to be same. so u can solve it.
    Last edited: Oct 24, 2011
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