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Pulley System with 2 Masses

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A 8.8 kg watermelon and a 7.7 kg pumpkin are attached to each other via a cord that wraps over a pulley, as shown. Friction is negligible everywhere in this system. Find the accelerations of the pumpkin and the watermelon. Specify magnitude and direction.

    [​IMG]

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    I have tried a few things but can't seem to figure out what I am doing wrong.

    First, I am having a hard time remembering trigonometry that may be helpful in determining angles, but I think I have remembered enough to be able to do this correctly.

    For the pumpkin I have
    mg=75.24N
    Fx=75.24sin37=45.41

    For the watermelon I have
    mg=86.24N
    Fx=86.24cos60=43.12

    Assuming the pumpkin is going in the negative x direction I get
    Sum of Fx=-2.29

    And using f=ma I get
    2.29=8.8a (for the for the watermelon)
    2.29=7.7a (for the pumpkin)

    aw=.260m/s^2 up the ramp
    ap=.297m/s^ down the ramp

    The directions are correct but the values are not. I have also tried with the angles of 53 and 30 in case I had not found those correctly but that did not come out correct either. Am I not able to combine the x direction forces because they are at different angles? Or is it something else I am missing? Thank you for any help.
     
  2. jcsd
  3. Oct 2, 2007 #2

    Dick

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    The magnitude of the accelerations of the pumpkin and the watermelon can't be different. They are tied together! You've forgotten to include a tension force from the cord in your force balance eqns.
     
  4. Oct 2, 2007 #3

    learningphysics

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    What direction is x supposed to be? Along the inclines? Are you calculating the component of gravity along the incline for Fx? I'm not sure how you get sin37...
     
  5. Oct 2, 2007 #4
    That makes sense...for some reason I was thinking because the watermelon had a greater m the a would have to be smaller, not thinking about differing forces...

    Wouldn't T for the pumpkin just be the x component of the watermelon force and vice versa though? That gives me the same value of -2.29 assuming the same axis directions.

    Yes, I put the x-axis along the incline in both cases. I got the 37 by comparing the triangle formed by the components of gravity to the triangle formed by each incline and I think that is the angle it matched up with (that is where I thought my trigonometry, or maybe just logic lol, may have been wrong).
     
  6. Oct 2, 2007 #5

    learningphysics

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    No it isn't. The acceleration of the two masses are the same magnitude wise. There is a shortcut to solving the problem, but I'd advise separating it into two parts as Dick is describing... ie examine the pumpkin alone and watermelon alone and get the force equations separately...

    You'll get 2 equations with 2 unknowns (T and a).

    The component of gravity along the incline for the first mass is m_pumpkin*gsin(53). For the second it's m_watermelon*gsin(30)
     
    Last edited: Oct 2, 2007
  7. Oct 2, 2007 #6
    I have used those values also so I guess I really have no idea how to find the angle correctly...

    For my two equations I get:

    T=m_pumpkin*gsin53+m_pumpkin*g*a

    and

    T=m_watermelon*gsin30+m_watermelon*g*a
     
  8. Oct 2, 2007 #7

    Dick

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    You are getting there! But you have to decide which direction is positive, down the pumpkin slope or down the watermelon slope? Once you've made that decision put signs on the forces and accelerations consistently.
     
  9. Oct 3, 2007 #8
    Ok I am having problems with this.

    If down the pumpkin slope is positive, then

    -T=-m_pumpkin*gsin53+m_pumpkin*g*a

    and for the watermelon

    T=m_watermelon*gsin30+m_watermelon*g*a

    I don't think I understand how the direction on one slope relates to the direction on the other. Do I think of positive values as all in the same direction (in this case to the left)?
     
  10. Oct 3, 2007 #9

    Dick

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    You are getting there. Put all of the forces on one side. If you choose positive to left then I would say +mg_pump-T=ma and -mg_water+T=ma. I've put plus on everything acting to the left and minus on everything acting to the right. If a turns out to be negative then that means the acceleration is really to the right.
     
  11. Oct 3, 2007 #10
    Ok I think I can figure this out...but solving both equations for T and then substituting it into the other equation gave me a=.106m/s^2 which was incorrect.

    Do I need to solve one in terms of a and substitute that way?
     
  12. Oct 3, 2007 #11

    Dick

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    Thats not what I get. Maybe you should show us exactly what equations you are solving. BTW why are you writing things like m*g*a? F=ma, right? Not mga.
     
  13. Oct 3, 2007 #12
    Ahh geez...

    I had been doing that the whole time and I never realized it, thank you for pointing that out. That makes things work out.

    Here are the equations I ended up with, I actually set positive x up the incline on the pumpkin side and down the incline on the watermelon side and I think it still worked out.

    Pumpkin
    T=ma+mgsin53

    Watermelon
    T=-ma+mgsin30

    so,
    7.7a+60.27=-8.8a+43.12
    16.5a=-17.15
    a=1.04m/s^2 with the pumpkin going down the incline and watermelon going up.

    I know this will probably make me feel worse about spending as long as I did on this problem, but what was the shortcut I could have used?

    Thank you guys a bunch for the help :)
     
  14. Oct 3, 2007 #13

    Dick

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    I don't think there is any shortcut. But now that you understand how to do the problem I'm sure it will go a lot faster in the future.
     
  15. Oct 3, 2007 #14

    learningphysics

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    By shortcut I was mainly just referring to the formula:

    [tex]\frac{m_{pumpkin}gsin(53)- m_{watermelon}gsin(30)}{m_{pumpkin} + m_{watermelon}}[/tex]

    This is just a variation on the standard atwood machine.

    I mainly mention this because when the situation gets more complex like on the atwood machines on this thread:

    https://www.physicsforums.com/showthread.php?t=187898&highlight=atwood+machines

    then it would be impractical to break it down the way it was done here.

    But I certainly advise doing the problem from the basics when possible, especially if one is encountering these concepts for the first time.
     
    Last edited: Oct 3, 2007
  16. Oct 3, 2007 #15
    Ah I see.

    I'm glad I was able to make it through (eventually) doing it the basic way, but I do see how that equation would be helpful for those more complex problems.

    Thanks again to both of you for the help.
     
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