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Newton's Forces Question

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Two crates, of mass 80 kg and 210 kg, are in contact and at rest on a horizontal surface (Fig. 5-26). A 750 N force is exerted on the 80 kg crate. The coefficient of kinetic friction is 0.21.


    (a) Calculate the acceleration of the system.
    .53 m/s2 (to the right) [Note: This Answer is Correct]

    (b) Calculate the force that each crate exerts on the other.
    _________ N
    (c) With the crates reversed, calculate the force that each crate exerts on the other .
    _________ N

    2. Relevant equations

    Ff= Mk(Normal Force)

    3. The attempt at a solution

    I did A, and I got .53, but I can't get B. I guess I just don't understand how to do it.

    I tried to calculate the force that Crate 1 exerts on Crate 2. I tried this but I got the wrong answer it says:




    It says that I'm wrong. :(

    I also can't get C...I don't even understand how to begin that one though.
    Last edited: Oct 17, 2007
  2. jcsd
  3. Oct 17, 2007 #2


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    It's Fx(net) = ma, but you have missed the applied 750N force in your equation which acts right, along with the crate force of B on A, as well as the friction force you correctly identified, which both act left. You should look also at a FBD of the second block. It has only 2 forces acting horizontally, so it's a bit easier to solve. For part c, reverse the blocks and do the same procedure.
  4. Oct 17, 2007 #3
    What is the crate force of B on A? :-/
  5. Oct 17, 2007 #4


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    That is the horizontal contact force of B on A(F_BA) that you are asked to solve, that acts perpendicular to the edge of the crate pointing left. Have you drawn a free body diagram of the left crate? Try drawing one also of the right crate. In this FBD of crate 2, you'll have the friction force acting left, and the horizontal contact force of A on B (F_AB)acting right. F_BA and F_AB are related by Newton's 3rd law. Note the acceleration of each block is the same as the accelertaion of the system.
  6. Oct 17, 2007 #5
    I'm terribly sorry but I still don't get it.

    <--Ff1--80 KG---750N-->

    weight 1

    <--Ff2--210 KG---750N--->

    weight 2

    Is that free body diagram correct? Are you saying that Ff2=750??? Because if you use the equation Ff=Mk(Fn)=432.12, not 750.
  7. Oct 17, 2007 #6
    how did u get A?
  8. Oct 17, 2007 #7
    A was simpler, for me at least:

    The total weight of the system is 290 kg. So m=290. a is our variable.





    All help is greatly appreciated on parts b/c.
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