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Solving friction problems

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Two trunks sit side by side on the floor. The larger trunk (52kg) is to the left of the smaller trunk (34 kg). A person pushes on the larger trunk horizontally toward the right. The coefficient of static friction between the trunks and floor is 0.35. Calculate the force the larger trunk exerts on the smaller trunk?

    mass of larger trunk = 52kg
    mass of smaller trunk = 34kg
    uk = 0.35

    2. Relevant equations

    uf = Ff/FN
    Fnet = sum of all forces

    3. The attempt at a solution

    FSmax = usFN
    FSmax = (0.35)(86kg)(9.8N/kg)
    FSmax = 290N

    FN = mg
    FN = (52kg)(9.8N/kg)
    FN = 509.6N

    Not quite sure if doing right steps or what to do next?
     
  2. jcsd
  3. Oct 24, 2015 #2
    Are you sure you've given us the problem? It seems something is missing.
     
  4. Oct 24, 2015 #3
    The magnitude of the maximum force the person can exert without moving either trunk is 290N! And that's basically it!
     
  5. Oct 24, 2015 #4
    Agreed, but nowhere in the statement of the problem is it mentioned that the maximum force is being applied. Neither do they state that the trunks aren't moving.

    Anyway, if you make those assumptions the next step is to look at the horizontal force applied to the 34 kg trunk. Or to the horizontal forces applied to the 52 kg trunk. Drawing the free body diagrams is a big help.
     
  6. Oct 24, 2015 #5
    I'm pretty sure the trunks are not moving meaning acceleration would be zero and so would Fnet? And I drew the FBD and just wondering would FT and FF cancel each other out?
     
  7. Oct 24, 2015 #6
    What are FT and FF?
     
  8. Oct 24, 2015 #7
    I think I made an error on my FBD, because I have yet to find FT but would FF be 290N?
    And I'm looking for FT right? And I'm just wondering if FT equals FSmax?
     
  9. Oct 24, 2015 #8
    What I meant was, how are you defining FT and FF?
     
  10. Oct 24, 2015 #9
    I think I figured it out

    Fsmax = usF N
    Fsmax = (0.35)(34kg)(9.8N/kg)
    Fsmax = 120N
     
  11. Oct 24, 2015 #10

    haruspex

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    Right, though as E=mc2 noted you have to fill in two missing pieces of information to arrive at that: that the trunks are not moving, but if the force between the trunks were any greater they would be. In the question as given, the pusher could be just breathing heavily against the larger trunk.
     
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