1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

One block on another

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A 104.0-kg block is on a horizontal frictionless surface, on top of it is a second smaller block of 68.0-kg. A horizontal force F of 613.4 N is applied to the upper block which accelerates with 5.84 m/s2. 1. Find the coefficient of kinetic friction between the two blocks.
    2. Find the acceleration of the lower block while the two blocks are in contact.


    2. Relevant equations
    F=ma, fk=(coeff kinetic)* F_Normal


    3. The attempt at a solution
    I found the coefficient of kinetic friction by taking the horizontal force (613.4N) and subtracting from it the force needed to move the block which is F=ma (68.0*5.84). That gave me a fk of 216.28N which I then divided by the normal force (mg= 666.4) to get the coefficient.

    I am having trouble finding the acceleration of the bottom block, I thought doing F=ma
    (613.3-216.28)=(104+68)a would give me the acceleration but it doesn't. Can someone give me a hint as in what to do because I can't figure it out.
     
  2. jcsd
  3. Oct 7, 2009 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In part 1, you looked at the forces acting on the top block. In part 2, you need to draw a free body diagram of the bottom block, and examine the forces acting on the bottom block.
     
  4. Oct 7, 2009 #3
    Ok, so I add the two masses together (i assume) which gives me 172 kg all together. The horizontal force is still 613.4 N so i set up the equation F=ma 613.4N=(172kg)*a which then gives me an acceleration of 3.566 m/s^2 but that's incorrect.
     
  5. Oct 9, 2009 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, when you take a free body diagram of the lower block, you 'isolate' it by itself and examine the forces acting on it. The 613.4 N force doesn't show up in such a diagram. What does show up is, amongst other forces, is the horizontal friction force from the top block. Use Newton 3 and 2 solve for the acceleration of the lower block. Make sure to use the correct value for the mass when using Newton 2. You shouldn't be adding the 2 masses together using this free body diagram approach.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: One block on another
Loading...