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PLEASE HELP! inclined pulley 2 masses

  1. Dec 30, 2008 #1
    1. The problem statement, all variables and given/known data

    This is a problem with two masses, a 24kg block sliding on an 11kg block. All surfaces are frictionless. Find the magnitude and acceleration of each block and the tension in the string that connects the blocks. The incline is 20 degrees. * this is not an atwoods machine...the blocks are on top of each other, connected by a massless pulley

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 30, 2008 #2

    rl.bhat

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    Homework Helper

    You can solve this problem just like Atwood's machine. But you have to replace g by gsin(20). Try.
     
  4. Dec 30, 2008 #3
    ok..so how would you start then? I tried it by using F=ma using the system, so i used the total mass and the net force( which is i think 117.31 using both the Fw parallels to the plane for each block). I thinkit may ybe different because its two masses on top of each other..not one hanging
     
  5. Dec 31, 2008 #4

    rl.bhat

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    Lower block is moving down along the incline plane with the upper block, even though the upper block is moving in the opposite direction. If T is the tension in the string, the acceleration of the lower block is [( 24 + 11 )g*sin20 - T]/35 and the acceleration of the upper block is [ T - 11gsin20]/11. Since they are connected by a single string, their accelerations must be equal. Equate them and solve for T and hence find acceleration.
     
  6. Dec 31, 2008 #5
    ok..so i tried doing that and solving for T, but the website (webassign) still marks it wrong. Also, the 24kg block is on top of the 11kg, so wouldnt the acceleration of the upper block be [T-24gsin20]/24 ?? Doing it this way, i got 89.26N as T. I also tried it your way and it marked it wrong.
     
  7. Dec 31, 2008 #6

    Doc Al

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    Staff: Mentor

    Treat this like any other Atwood's machine problem. The only difference is that all relevant forces are parallel to the incline instead of vertical. Set up Newton's 2nd law for each mass, giving you two equations. Solve for the two unknowns.
     
  8. Dec 31, 2008 #7
    ok, i did that and webassign marked the answer wrong. I followed rl.bhat's help by setting the two equations equal to each other because the accelerations are equal, and i solved for T. Then i plugged that in to find a, but i found that the a's in the two different equations were different. Is there something special because there is a larger mass on top of a smaller mass? Does it move up the incline, not down? Maybe there's something wrong with my math because of the directional negatives
     
  9. Dec 31, 2008 #8

    Doc Al

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    Show exactly what you did. What were your two equations?
     
  10. Dec 31, 2008 #9
    never mind...i got the answer by adding up the two net equations for each block and then solving for a. I then plugged that answer into the net force equation for one block to find T..but what's weird is that i got a different answer for T when i plugged it into the 11kg block (the wrong answer) but got the right answer when i solved for T in the 24kg net force equation. Why do you think this is? Shouldnt T be the same in each?
     
  11. Dec 31, 2008 #10

    Doc Al

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    Of course T must be the same in both equations. That fact is used to solve for the acceleration. Unless I see the equations that you used, I don't really know what you did.
     
  12. Dec 31, 2008 #11
    for the 24kg i did T-Fwsin20=24a, for the 11 kg Fwsin20-T=11a, then i added them together to get the equation, 107.8sin20-235.2sin20=35a. Then i got -1.245 for a. Then i attempted to plug a into each of the original equations to solve for T, but got two different values (the correct answer is 50.56N which i only got when i plugged in a into the equation for the 24kg block). Maybe my math is wrong?
     
  13. Dec 31, 2008 #12

    Doc Al

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    Those equations and your answers are fine. And you get the same answer for T using either equation. Try it again. (Careful with signs.)
     
  14. Dec 31, 2008 #13
    ohh ok i got it...thank you!!
     
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