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Homework Help: Need help with problem

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Hi, I have a problem here and I need some help on it. The problem is this:


    block m[tex]_{1}[/tex] weighs 875 N. The coefficient of static friction between the block and the table is 0.24 and the angle theta is 25.0[tex]\circ[/tex]. Find the maximum weight of block m[tex]_{2}[/tex] for which block m[tex]_{1}[/tex] will remain at rest.

    2. The attempt at a solution

    So far I've drawn an FBD for m[tex]_{1}[/tex], m[tex]_{2}[/tex], and the knot that connects them.

    For the block m[tex]_{1}[/tex], I have T[tex]_{1}[/tex] = [tex]M[/tex][tex]_{s}[/tex]N in the x-direction, and N = m[tex]_{1}[/tex]g for the y-direction.

    For block m[tex]_{2}[/tex], I have T[tex]_{2}[/tex] = m[tex]_{2}[/tex]g in the y-direction.

    For the knot, I have T[tex]_{1}[/tex] = T[tex]_{3}[/tex]*cos(25) in the x-direction and T[tex]_{2}[/tex] = T[tex]_{3}[/tex]*sin(25) in the y-direction.

    Since T[tex]_{1}[/tex] = T[tex]_{3}[/tex]*cos(25), and T[tex]_{1}[/tex] = [tex]M[/tex][tex]_{s}[/tex]N, I set T[tex]_{3}[/tex]*cos(25) = [tex]M[/tex][tex]_{s}[/tex]N and solving for T[tex]_{3}[/tex], I get T[tex]_{3}[/tex] = ([tex]M[/tex][tex]_{s}[/tex]N)/(cos(25))

    In the knot, we have T[tex]_{2}[/tex] = T[tex]_{3}[/tex]*sin(25). I plug in the T[tex]_{3}[/tex] derived earlier and T[tex]_{2}[/tex] = m[tex]_{2}[/tex]g into this equation to get m[tex]_{2}[/tex]g = [tex]M[/tex][tex]_{s}[/tex]*m[tex]_{1}[/tex]g*tan(25)

    When I plug in numbers, the answer I get is about 98, and that just doesn't seem right to me. Did I do something wrong? Thanks for the help in advance.
  2. jcsd
  3. Oct 5, 2007 #2


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    Homework Helper

    I get the same answer as you. 97.9N. Why do you think it is the wrong answer?
  4. Oct 5, 2007 #3
    I've tried putting that in as the answer, but the computer says it isn't correct.
  5. Oct 5, 2007 #4


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    Homework Helper

    Did you put in 98 or 97.9? Maybe they want the mass and not the weight? I'm not sure...
  6. Oct 5, 2007 #5
    When I put it in, it wants the units as newtons, so I'm sure it's weight. But now that I have a second opinion, I'll just argue with the professor if it's wrong =]. Thanks!
  7. Oct 5, 2007 #6
    Did you neglect the static friction force?

    which would be 210N.. So whatever Tension you assigned to the rope being pulled horizontally would be a maximum of 210N
  8. Oct 5, 2007 #7
    Yes it seems you did cause I also get 97.9N as the answer....

    If it helps I am also a Physics teacher for the sake of your arguement.
  9. Oct 5, 2007 #8
    I did the problem, and I also got 97.9 N. Just to let you feel more secure about it :)
  10. Oct 5, 2007 #9
    Alright! Thanks everyone!
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