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Finding the mass of a block

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Block A with a mass of 10 kg rests on a 35degree incline. The coefficient of static friction is 0.40. An attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. The largest mass mb, attached to the dangling end, for which A remains at rest is....?


    2. Relevant equations
    F = ma
    Fs = us*Fn



    3. The attempt at a solution
    Fs = .4*10*9.8*cos35 = 32.1N

    That's all I've got so far. :rofl:

    Not sure why I'm struggling so hard on this problem.
     
  2. jcsd
  3. Nov 10, 2008 #2

    tiny-tim

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    Hi hansel13! :smile:

    Have you drawn an fbd for the block?

    There's the friction (which you've caclulated), the tension, the normal force, and the weight.

    So what is the tension? :smile:
     
  4. Nov 11, 2008 #3
    lets calls the mass of block B, B for now.

    I think I figured it out, so:
    T = Fk-mg*sin30
    = 32.1N -10kg * 9.8m/s2*sin30 = 24.1


    And we know that T = B*g

    So B = T/g 24.1/9.8 = 2.46 kg

    I'm pretty sure that's right, is it?

    Thanks for the help! Lots of times I get lost and forget what I'm actually looking for, when you mentioned that I needed to find Tension I just drew it up and figured it out.
     
  5. Nov 12, 2008 #4
    Hopefully it is haha..
     
  6. Nov 13, 2008 #5
    If someone could let me know if this is right or not, I'd really appreciate it. I'm not too confident in my physics skills yet and this is an extra 5 points to my exam correct, so yeah...
     
  7. Nov 13, 2008 #6

    tiny-tim

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    Hi hansel13! :smile:

    Yes, that's more-or-less right, except …

    i] isn't it 35º? :rolleyes:

    ii] shouldn't you add the weight force to the friction force?
     
  8. Nov 13, 2008 #7
    ah, Right!

    T = Fk-mg*sin30
    = 32.1N + 10kg * 9.8m/s2*sin35 = 88.2


    And we know that T = B*g

    So B = T/g 88.2/9.8 = 9kg

    Thanks! Dumb mistakes on my part..
     
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