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Two Masses Connected by Pulley; find acceleration

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Mass m1 = 11.9 kg is on a horizontal surface. Mass m2 = 8.80 kg hangs freely on a rope which is attached to the first mass. The coefficient of static friction between m1 and the horizontal surface is μs = 0.526, while the coefficient of kinetic friction is μk = 0.168.
    (There is a picture with M1 on top of a ...lets say table..and the m2 box is on the side of the table with a pulley on the corner)
    If the system is in motion with m1 moving to the left, then what will be the magnitude of the system's acceleration? Consider the pulley to be massless and frictionless.

    2. Relevant equations
    Extra notes from professor:
    The frictional force will always oppose the direction of relative motion (or impending motion). If the block on the horizontal surface is moving to the right, then the frictional force is to the left. If the block on the horizontal surface is moving to the left, then the frictional force is to the right. However, you should see from a force diagram of the block on the horizontal plane, the tension in the string is to the right in both cases (tensions can only pull).

    3. The attempt at a solution
    I've tried different things by examples my professor has given me, however those included angles so I'm not sure how to do this one.
  2. jcsd
  3. Sep 20, 2011 #2


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    It appears that the hanging mass is on the right side of the table and moving up, and the block on the table is moving left, correct? Draw a free body diagram of each block, identify the forces acting (with the help of your professor's hints), and use newton 2 to solve for the acceleration . You have to solve 2 equations with 2 unknowns.
  4. Sep 20, 2011 #3
    Newton 2 being?
    SigmaF = ma for each?
  5. Sep 20, 2011 #4
    Yes. Newton's 2nd law states that.....

    [itex]\Sigma[/itex]F = ma
  6. Sep 21, 2011 #5
    Okay. So...
    For M1: there is natural, gravity, and tension ?
    For M2: there is natural, gravity, weight, and tension?
  7. Sep 21, 2011 #6
    M1 T - m1*g = m1*a
    M2 T + m2*g = m2*a
  8. Sep 21, 2011 #7
    M1: Normal (upward), Weight (downward), Tension (rightward), Friction (rightward)
    M2: Tension (upward), Weight (downward)

    M1's equation

    [itex]\Sigma[/itex]F = m[itex]_{1}[/itex]g - m[itex]_{1}[/itex]g - T[itex]_{1,2}[/itex] - [itex]\mu[/itex]m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a

    M2's equation

    [itex]\Sigma[/itex]F = -m[itex]_{2}[/itex]g + T[itex]_{2,1}[/itex]= m[itex]_{2}[/itex]a

  9. Sep 21, 2011 #8
    Okay. And w = mg F = ma now what?
  10. Sep 21, 2011 #9
    Now, you add your 2 equations together.....

    [itex]\Sigma[/itex]F = -m[itex]_{2}[/itex]g - [itex]\mu[/itex]m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a + m[itex]_{2}[/itex]a

    Solve for a.
  11. Sep 21, 2011 #10
    okay. i guess i just dont understand how to get the equations.
  12. Sep 21, 2011 #11
    If you want, I can try to explain how to get these equations. :smile:
  13. Sep 21, 2011 #12
    In order to come up with these equations, you have to first figure out what forces are acting upon the objects in question.

    The four basic ones when dealing with pulleys are:
    1. Weight (otherwise known as gravitational)
    2. Normal (otherwise known as natural in some cases)
    3. Tension
    4. Friction.

    As you already know, weight forces tend to be in the pointed downwards. Normal forces tend to be pointed upwards.

    Friction forces and Tension forces vary based on the circumstances.

    So, let's consider this problem.

    We have a weight force pointing downward and a tension force pointing upward on M2.

    On M1, we have a weight force pointing down, a tension force pointing to the right, a friction force pointing to the right, and a normal force pointing upwards.

    Weight force = -mg
    T force for M1 = T force for M2
    Friction force = u(N)
    Normal force = mg

    Now, when we sum our forces for the M1 object....

    [itex]\Sigma[/itex]F = ma

    The sum is just all of the above forces added together and set equal to ma.

    [itex]\Sigma[/itex]F = -mg + mg + (-T) - u(N) = ma

    Now, T is negative because it is pointing opposite of the direction of motion, and friction is negative for the same reason.

    Now, for M2....

    We have a W force = -mg and a T force.

    So, [itex]\Sigma[/itex]F = T - mg = ma

    T is positive because it is pointing upward and W is negative because it is pointing downward.

    And that's a specific way of how to get the equations for this problem. :smile: Hopefully it cleared things up a bit.
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