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Inclined plane pulley, help

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data

    M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (6.50kg) accelerates downwards at 3.27m/s2, that the angle theta is 20.0o, and that muk is 0.340.


    2. Relevant equations


    3. The attempt at a solution

    I drew my free body diagram for mass M1 and M2, and I also know that the acceleration is the same for both masses. Also, I know that the tension in the string should also be the same for the two masses. From there, I am kinda lost. Any help...
  2. jcsd
  3. Oct 6, 2007 #2
    can you tell us what you got for newton's second laws on each of the 2 bodies?
    from there you can combine and solve for m1.
  4. Oct 6, 2007 #3
    for M1 there is going to be a friction force parallel to the plane opposing the motion up the plane. There is a normal force perpendicular to the plane and mg going straight down. Then there is the force from M2 that causes the M1 to move up the ramp.
  5. Oct 6, 2007 #4
    Kay but write it algebraically and don't plug in any of the numbers at first and that way you can get a formula for m2 which will depend on the angle/m1 and other things.
  6. Oct 6, 2007 #5
    that is the part I am struggling with. This is my first time taking physics.
  7. Oct 6, 2007 #6
    If you wait a little bit, I'll try my best attempt at the problem
  8. Oct 6, 2007 #7
    alright sounds good
  9. Oct 6, 2007 #8
    ok well then let's find all of the forces first:

    friction force opposing the motion up the plane
    Normal force
    mg going straight down (best to split this into it's components)

    ma downward

    What other force is missing? From the string connecting the blocks right?
  10. Oct 6, 2007 #9
    For M2 there should be a tension force upwards. For M1 there should also be a tension force up the ramp.
  11. Oct 6, 2007 #10
    right, can you attempt to write it algebraically? you know all the forces and if say the + axis is going up and to the right you can determine whether that force is aiding the net force or resisting it.

    I'll start ya off, for m2 you will have something like:

    mg is - since we assume that up is positive, you could say down is positive but you have to keep using that assumption always, you can't say for m2 down is + but for m1 down is -.
  12. Oct 6, 2007 #11
    The change in the forces in the x and y direction for block one is
    F_x = T - F_f = (m_1)a (Friction force = F_f)
    F_y = F_n - (m_1)gcos* = 0 (normal force = F_n)
    and for block two is
    F_x: (m_2)g - T = (m_2)a
    There is no y component on the second block
    To help you understand why, rotate m_2 so that the ropes make a straigt line. Then mg is to the right. Realistically it's not, but it helps you understand how the T's are connected

    From there, you should be able to substitute equations to find m_1
    Last edited: Oct 6, 2007
  13. Oct 6, 2007 #12
    tension is aiding the net force, and friction is opposing it. It makes sense what you wrote for M2 but I don't know how to do it for M1.
  14. Oct 6, 2007 #13
    If you need help (don't look until you try yourself):
    T = F_f + (m_1)a = uf_n + (m_1)a
    f_n = (m_1)gcos*
    T = u(m_1)gcos* + (m_1)a

    (m_2)g - (m_2)a = T
    (m_2) = T/(g-a)
    T = (m_2)(g-a)

    u(m_1)gcos* + (m_1)a = (m_2)(g-a)
    (m_1) = (m_2)(g-a)/(ugcos* + a)
    sorry, had to edit it a bit
    Last edited: Oct 6, 2007
  15. Oct 6, 2007 #14
    well what forces are helping the block go up? what forces are opposing the motion upwards?

    just the sum of those helping are added, opposing are subtracted will = m2a.
  16. Oct 6, 2007 #15
    I got 13.56 kg as my answer
    Which, looking over, is unrealistic since M1 needs to be lighter than M2
    I'm still looking over my work, so I wont give up~
    Last edited: Oct 6, 2007
  17. Oct 6, 2007 #16
    11.97 is not right. Not sure where we went wrong.
  18. Oct 6, 2007 #17
    I got 13.56 kg also, but it says that it is incorrect. I used the (m_1) = (m_2)(g-a)/(ugcos*)
  19. Oct 6, 2007 #18
    I'm still working on it, after I edited my work I got 6.63 now, but that still doesn't seem right. I'm getting closer. so I'll keep trying.
  20. Oct 6, 2007 #19
    I'm in the middle of editing this.. so it will change, just leaving it hear so I have everything in order~

    F_x = T - F_f = (m_1)a (Friction force = F_f)
    F_y = F_n - (m_1)gcos* = 0 (normal force = F_n)
    and for block two is
    F_x: (m_2)g - T = (m_2)a

    T = F_f + (m_1)a = uf_n + (m_1)a
    f_n = (m_1)gcos*
    T = u(m_1)gcos* + (m_1)a

    (m_2)g - (m_2)a = T
    (m_2) = T/(g-a)
    T = (m_2)(g-a)

    u(m_1)gcos* + (m_1)a = (m_2)(g-a)
    (m_1) = (m_2)(g-a)/(ugcos* + a)

    Using that, I had gotten 6.63 kg.. but it should be less still.
  21. Oct 6, 2007 #20
    Yeah that still doesn't make sense because M1 must be less than M2. Thanks
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