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Pulley question - acceleration of an object across a horizontal surface

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    The figure shows two objects of masses m(1) and m(2). The horizontal surface allows for frictionless motion. The string ties to the two objects is massless and passes over a massless pulley that rotates without friction. If m(1)=5.63 kg and m(2)=1.57 kg, what is the magnitude of the acceleration of the sliding object?

    http://session.masteringphysics.com/problemAsset/1073792/5/12.P71.jpg

    We know that:
    - tension m(1) = tension m(2) because they're connected
    - mass of m(1), therefore we know the weight because W=mg
    - mass of m(2) and we can calculate the weight too
    - acceleration is a non-zero value, so Fnet=ma



    2. Relevant equations
    Fnet=ma



    3. The attempt at a solution
    - For the FBD for m(1), the larger object, wouldn't Fnet(x direction) = T= Ma, and Fnet (y direction) = 0 because it's on a flat surface, meaning that the normal and weight cancel out? SO the first equation would be Fnet(x)T=Ma (1)
    - For the FBD for m(2), the smaller object, wouldn't Fnet(x direction)=0 since the object is just suspended in the air, and Fnet(y direction)=T-mg... so the 2nd equation would be Fnet(y)T=ma+mg (2)

    (I equated Fnet in each axis to mass object * acceleration because they're both moving in that direction)

    And since the Tension is the same, you could equate them to each other.. I equated
    m(1)a=m(2)a+m(2)g, then isolated for a, but got the wrong answer. What am I doing wrong?!
     
    Last edited: May 25, 2009
  2. jcsd
  3. May 25, 2009 #2
    For your smaller object mg is bigger than T otherwise it wouldn't accelerate.
    Fnet=mg-T.You can take it from there.
     
  4. May 25, 2009 #3
    Thanks, that helped a lot. So it turns out that my methods of solving the problem were correct, its just that I didn't realize that W(2) > T... which is what allowed it to accelerate in the direction of W(2). Downwards.
     
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