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Homework Help: 2 trucks pulling a third truck by the means of a pulley

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Two trucks tow a third one by means of inextensible ropes and a pulley attached to them (fig. 1). The accelerations of the two trucks are a1 and a2. What is the acceleration of the third truck that is being towed?
    (fig.1) is in this link (I'm sorry, I don't know how to post images on websites):
    http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_AS_2007_QP.pdf

    m=mass of truck
    a1,a2 respective accelerations
    a=the shared acceleration (whole system)
    T= Tension

    2. Relevant equations
    F = ma
    Free body diagrams

    3. The attempt at a solution
    Although the problem didn't specify, I'm assuming the masses of the trucks are equal
    Since the total acceleration of the two trucks must be the same,

    2ma = ma1-ma2
    Thus, a= (a1-a2)/2
    Furthermore, ma =T - ma2 for the second truck, so

    T = ma + ma2 -> T = m(a1+a2)/2

    The force of the tension on the pulley and 3rd truck = 2T

    And total force on the pulley, which pulls the third truck = ma

    Thus ma = 2T -> a = a1+a2, but the answer key says it's a = (a1+a2)/2
     
  2. jcsd
  3. Jan 16, 2016 #2

    TSny

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    Hello. Welcome to PF!

    The accelerations of trucks 1 and 2 are not necessarily the same. This is a kinematics problem. You do not need to work with forces.

    For problems like this, it is often helpful to use the fact that the total length of a rope must remain constant. Consider introducing distances from a fixed line ("datum") as shown below. Can you relate the distances x1, x2, and xp to the length of the rope connecting trucks 1 and 2?
     

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  4. Jan 16, 2016 #3
    So that'll be length = x1 + x2 -2xp ?
     
  5. Jan 16, 2016 #4
    Will it be then length of rope = x1 + x2 - 2xp = constant

    And perhaps take the double derivative of that?
     
  6. Jan 16, 2016 #5

    TSny

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    Yes!
     
  7. Jan 16, 2016 #6
    Ohh I see! That's a really elegant way to do that. Do you think it's possible with forces, just for the fun of it? Or will it be difficult to do that?
     
  8. Jan 16, 2016 #7

    TSny

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    The kinematic constraint is what determines the relation between the accelerations, not the forces. But, you can then use the relation between the accelerations and F = ma to relate various forces in the problem.
     
  9. Jan 16, 2016 #8
    I see, so the accelerations are dependent on the constraint and not always on the forces situations such as this one. So for example if I were to determine the tension in the string, since 2T = ma where a must be the acceleration of the pulley, I can determine T = m(a1+a2)/4 and etc. Am I on the right track?
     
  10. Jan 16, 2016 #9

    TSny

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    Yes. The details would depend on what information you are given and what you are trying to determine.
     
  11. Jan 16, 2016 #10
    Ahh I see, thank you so much! Im glad I came here for help ^^
     
  12. Jan 16, 2016 #11

    TSny

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    OK. Glad I could help.
     
  13. Feb 26, 2018 #12
    How do you take the double derivate of that and come to the answer?
     
  14. Feb 26, 2018 #13

    TSny

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    Welcome to PF!

    Please write out the specific equation that you wish to differentiate.
     
  15. Feb 26, 2018 #14
    The one used to answer the question above...so x1+x2-2xp.I get that the length of the rope is x1+x2-2xp but I don't understand how you differentiated it and how that led to the answer.
     
  16. Feb 26, 2018 #15

    TSny

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    So the equation is L = x1 + x2 - 2xp, where L is the length of the rope. What do you get if you take one time derivative of each side?
     
  17. Feb 26, 2018 #16
    I don't know.
     
  18. Feb 26, 2018 #17

    TSny

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    What does the time derivative of x1 represent?
     
  19. Feb 26, 2018 #18
    Velocity?
     
  20. Feb 26, 2018 #19

    TSny

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    Yes. So, you could write the time derivative of x1 as v1.
     
  21. Feb 26, 2018 #20
    Where are you going with this?

    So, x1+x2-2xp=l
    v1+v2-2vp=vl?
    a1+a2-2ap=al?
    a1+a2-al=2ap?
    (a1+a2-al)/2=ap?
     
  22. Feb 26, 2018 #21

    TSny

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    Hopefully to some insight :oldsmile:

    What does vl represent? The ropes are given to be "inextensible".
    What does this imply about the right hand side of your equation x1+x2-2xp=l?
     
  23. Feb 26, 2018 #22
    vl=0 there is no change is l
    you could rearrange to get vp=(v1+v2)/2
    I guess if you do the second derivative you get ap=(a1+a2)/2
    Am I right?
     
  24. Feb 26, 2018 #23

    TSny

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    Yes. Good work.
     
  25. Feb 26, 2018 #24
    OMD! I finally got there. Thanks again for your patience and quick reply! Is there a name for this technique using this 'datum' thing or is it just logic?
     
  26. Feb 26, 2018 #25

    TSny

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    I don't know of any name for the technique.
     
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