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Three Blocks Accelerating

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9.8 m/s2. Three blocks are sliding on a frictionless horizontal surface:

    Fℓ----->[Block1-1kg][Block2-7kg][Block3-4kg]<--Fr

    If Fℓ = 29 N and Fr = 3 N, what is the magnitude of the force exerted on the block with mass 7 kg by the block with mass 4 kg?

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    29 N - 3 N = 26 N (total force) 1kg + 7 kg + 4 kg = 12 kg (total mass)

    a= F/m = 26 N/12 kg = 2.166666667 m/s2 (acceleration of all the blocks in the right direction)

    masses of Block2 + Block 3 = 11 kg

    F = (11 kg)(2.166666667) = 23.83333333 N (THIS IS THE WRONG ANSWER!)

    What am I doing wrong?
     
  2. jcsd
  3. May 23, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You found the net force on the "Block2 + Block 3" system. Not sure why.

    Hint: What's the net force on Block 3? How can you use that result to answer the question asked in the problem?
     
  4. May 23, 2009 #3
    Net force on Block 3 = (4 kg)(2.166666667) = 8.666666668 N

    So, the force of 8.67 N is being exerted on Block 3 from the left, but if I am trying to find the force being exerted on Block 2 from Block 3 (from the right), then would I have to find the net force being exerted by Block 1 and 2? I'm not quite sure.
     
  5. May 23, 2009 #4
    OR:

    The net force on Block 3 + Fr = 8.67 N + 3 N = 11.67 N (force exerted on block 2 by block 3)

    Is that right?
     
  6. May 23, 2009 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    No, 8.67 N is the net force on block 3.

    Don't forget Newton's 3rd law: If you can figure out the force that block 2 exerts on block 3, then you automatically know the force that block 3 exerts on block 2.

    Yes. Here's how to think of it:
    Net Force on Block 3 = + 8.67 N (Positive, since it points right.)
    Force of Block 2 on Block 3 + Fr = + 8.67 N
    Force of Block 2 on Block 3 - 3N = + 8.67 N (Note that Fr is negative, since it points left.)
    Force of Block 2 on Block 3 = + 8.67 N + 3N = 11.67 N

    Thus the force of Block 3 on Block 2 = -11.67 N (from Newton's 3rd law); the magnitude is just 11.67 N.
     
  7. May 23, 2009 #6
    Thank you!
     
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