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A .5 kg block of cheese sits on a table (Static Friction q)

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data

    A .5 kg block of cheese sits on a level table. The coefficient of static friction is .60. Three strings are tied together in a knot at a point, K. One string is tied to the cheese, the other on a wall at a 30 degree angle, and in the middle of those two strings a mouse hangs from them. What is the maximum mass of the mouse for the cheese and mouse to remain in equilibrium.

    2. Relevant equations

    F = ma Fg = mg

    3. The attempt at a solution

    So, I calculated the Ff of the cheese, being 2.948 N.
    I know I need to calculate the Ft of the left string, and I think that equation would be for Ftx = (mass mouse)(-9.81) / tan30.

    I don't know how to calculate Fty, and I don't j=know how to finish the question up. The answer is .16kg.
     
  2. jcsd
  3. Oct 18, 2015 #2

    haruspex

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    I assume the string from the cheese to the knot is horizontal. You mention tension in "the string", but there are three strings. Please use distinct symbols for the different tensions. What statics equations do you have at the knot? At the cheese?
    By the way, seems to me you are almost there.
     
  4. Oct 18, 2015 #3
    Since you don't show the diagram, I would assume that the string attached to the cheese is parallel to the table.
    If that's the case then the vertical component of the tension in the string (at 30 deg) would have to support the mouse and
    the horizontal component of that string would provide the force needed to move the cheese.
    It looks like that answer given is slightly less than the mass calculated using this solution.
    Perhaps this is because the cheese could move using this value for m.
     
  5. Oct 18, 2015 #4
    Sorry, the cheese string is in fact parallel. The coefficient of friction between the cheese and the table is .60. I don't get how to find tension in the right string (the 30 degree one) without being given mass, because mass of the mouse determines the tension?
     
  6. Oct 18, 2015 #5

    haruspex

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    Draw the free body diagram for the knot. There are three forces on it. What two equations can you write down?
     
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