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Friction, impeding motion

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data
    The coefficients of friction are Us = .4 and Uk = .3 between all surfaces of contact. Determine the force P for which motion of the 30kg block is impendinging if cable AB is attatched as shown


    3. The attempt at a solution
    wups i forgot to lable AB but theres only 1 cable so the picture is clear

    im having trouble setting this one up. i dont know whether to start on the bottom block or the top. and im not sure if im setting up my forces correctly. heres what i have for the top block:

    sum Fy = N - (20)(9.8) = 0 ---> N = 196N
    sum Fx = Tab - Us(N) = 0 ---> Tab = Us(N) = .4(196) = 78.4N

    is that much correct? if it is then im not sure if im setting up the next part right. for the bottom block i have:

    sum Fy = N - (20)(9.8) - (30)(9.8) = 0 ----> N = 490N
    sum Fx = -P + Tab + (Us)(N) = 0 ---> P = 78.4 + (.4)(490) = 274.4N

    if i screwed up anything, its probably the horizontal forces. is Tab right or should it be (tab)(Us) ?

    Attached Files:

  2. jcsd
  3. Nov 29, 2008 #2


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    Homework Helper

    If you look at it as being equivalent to the 30 kg and the 20 kg taken twice contributing to friction that's 70*g*.4 which is 274.4 as you've found.

    I'd say you are ok.
  4. Nov 29, 2008 #3

    Doc Al

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    Staff: Mentor

    A couple of problems: For one, the cable only pulls on the top block, not the bottom block. So cable tension shouldn't appear explicitly in the analysis of the bottom block. Also, there are two friction forces acting on the bottom block.

    Nonetheless, your answer is good. :wink: (Since the cable tension is equal to the other friction force, you get the right answer.)
  5. Nov 29, 2008 #4
    hm. ok so what you're saying is i didnt really even need a freebody diagram for the top block, i could have just gone straight to the bottom and put (20kg)(9.8)(.4). ok thanks
  6. Nov 29, 2008 #5

    Doc Al

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    Staff: Mentor

    That's right. All you need is a freebody diagram for the bottom block, which would show two friction forces. (Of course, you'd need a FBD for the top block to calculate the normal force between the two blocks, if it wasn't obvious.)
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