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Acceleration of Two Objects on a Pulley

  1. Dec 29, 2008 #1
    1. The problem statement, all variables and given/known data

    The problem is as follows:

    A 10 kg object and a 20 kg object are hung on a pulley system. Find the acceleration of each object.

    3. The attempt at a solution

    I have already drawn the free-body diagrams, but I am confused as to where friction occurs in this problem. I have written out an equation for the first object - FNET=sigma F=T-W+a. I'm following Newton's Second Law, but how can I make what I have already written equal to ma if a is already in the first part of the equation? Also, how do I include friction, if it is necessary? Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 29, 2008 #2

    Doc Al

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    This doesn't quite make sense. Why are you adding the acceleration to the forces?
    But it shouldn't be in the first part of the equation.
     
  4. Dec 29, 2008 #3
    I added acceleration because I thought the first object would be accelerating upwards, and the second object would be accelerating downwards, due to the amount of their masses. I guess I misunderstood?
     
  5. Dec 29, 2008 #4

    Doc Al

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    The 10 kg mass will accelerate upwards, thus its acceleration would be represented by +a. Similarly, the downward acceleration of the 20 kg mass would be represented by -a. But you would never add acceleration to the forces. Acceleration and force are two different quantities with different units--they never add.

    Acceleration appears on the right hand side of Newton's 2nd law: ∑F = ma. That's where it belongs.
     
  6. Dec 29, 2008 #5
    Ok, I moved the acceleration to its proper location in the equation. I then had T-W=ma. After that, I changed my equation to a=(T-W)/m to solve for the acceleration. I know how to use w=mg to find the weight and the mass was a given, but I'm not sure how to calculate the tension. Where do I go from here?
     
  7. Dec 29, 2008 #6

    Doc Al

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    Analyze the forces on and apply Newton's 2nd law to each mass. You'll end up with two equations, which you can solve for the two unknowns. (Be careful with signs.)
     
  8. Dec 30, 2008 #7
    As always, draw a Free-Body Diagram of what is happening. This will allow you to apply: sum of forces = mass*acceleration.

    Note: This only works for constant mass. Keep in mind of this, Newton's Second Law is really:force = change of momentum with respect to time.

    Then you would apply this concept to both, so you'll be solving it for Block A, and Block B. As said above, you will have two equations. Subsequently, you can use some clever algebra to put one of the equations into the other and solve for the acceleration of one of the blocks.

    So look back at your free-body diagram and look at what's happening with block a. Write the equation for that. Disregard block a now, and look at block b and how it sees what's going on. Write an equation for that.

    Now you have two equations, and with these two equations there are common things such as: mass, tension etc. Use some algebra to get it into a form to plug it into the other equation, then when you've combined the equations you can now solve for acceleration of one of the blocks. Then do the same to find the other.
     
  9. Dec 30, 2008 #8
    Generally mass is constant :wink: Make sure that acceleration is constant too.
     
  10. Jan 1, 2009 #9
    I'm at the point where I have the two equations, a1=(T-W1)/m1 and -a2=(T-W2)/m2. I'm not seeing any factors that I could substitute into one of the equations. Am I missing anything in my equations?
     
  11. Jan 1, 2009 #10
    If the two objects are attached via a rope and strung about 1 pulley, what can you say about a1 and a2 ?
     
  12. Jan 1, 2009 #11
    I suppose that would mean that the two accelerations would be equal to each other, making a1 = -a2.

    (T-98N)/(10kg) = (-T+196N)/(-20kg) is what I come up with now. I have no idea how to figure out the tension. From what I currently have written, the accelerations equal -9.8 m/s squared. Is finding out the tension the last step I need to do? And how do I go about finding tension?
     
  13. Jan 1, 2009 #12

    Doc Al

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    OK.

    Since you rearranged each equation to have the acceleration in terms of tension, you'll have to first solve for the tension to find the acceleration. You have an equation with one variable. Solve for T.

    Better would have been to keep the equations like so:
    T - M1g = M1a
    M2g - T = M2a

    Then you could combine them to eliminate T and solve for a directly. (How would you combine them?)
     
  14. Jan 1, 2009 #13
    I tried adding T - M1g = M1a to M2g - T = M2a in an effort to eliminate T. My work is as follows:

    M2g-M1g = (M1+M2)a
    a = (M2g-M1g)/M1+M2
    (20kg)(9.8m/s2) - (10kg)(9.8m/s2) / 10kg + 20kg

    = 3.27m/s2

    Is this the correct answer?
     
  15. Jan 1, 2009 #14

    Doc Al

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    Looks good to me.
     
  16. Jan 1, 2009 #15
    Yay! Everyone, thank you so much for your help.
     
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