1. Not finding help here? Sign up for a free 30min 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!

Archived Dynamics- Two Body Problems

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Two blocks are stack on each other. The lower block has a mass of 6.8kg and μs=0.34 with the floor. The top block has a mass of 3.4kg and μs=0.65 between the two blocks. A force of 18N is being applied to the top box to the left. What maximum force can be applied to the lower block without the upper block slipping? What is the acceleration of the system?

    2. Relevant equations

    Fx= max

    3. The attempt at a solution
    I drew free body diagrams for the system, and started the 3 equations as the following;

    for mA,
    Fx= mAax
    T- μsη-mAgsinθ= mAax 1)


    Fy= maay
    η-magcosθ=0


    for mB,
    Fx= mBax
    mBg-T=mBax 2)


    I think I should add 1) and 2) to each other so I can get rid of T but I'm not sure about where I should go from here.
     
  2. jcsd
  3. May 1, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Since the OP has chosen an unusual set of variable names and not defined them, I cannot comment on the equations given.

    The first thing to check is the state of things with no force applied to the lower block. We find that the 18N is sufficient neither to make the top block slide on the lower block, nor to make both blocks slide on the table.
    Next, we must ask what the force applied to the lower block must do in order to make the blocks slide relative to each other. If it also makes nothing move then the 18N still won't. So the extra force must make the lower block accelerate so fast that the combination of the top block's inertia and the 18N force overcomes the static friction.

    |mAsA|<|FA+amA|, where the acceleration a is measured oppositely to the 18N force.
    Note that this has two solutions.
    To achieve this acceleration, a(mA+mB)=FB+/-(mA+mB)gμkB, where the +/- sign is opposite to the sign of FB.
    Since we are not told any coefficients of kinetic friction, we cannot go further without making assumptions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dynamics- Two Body Problems
  1. Two Body Problems (Replies: 3)

Loading...