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External forces

  1. Jan 6, 2010 #1
    1. A car of mass 1000kg is towing a trailer of mass 650kg and the two have an acceleration of 2,3m/s^2. Assuming that the only external force is the driving force between the wheels and the road, calculate:
    1: the value of this force
    2: the tension in the coupling between the car and the trailer.

    I'm a little unsure about this question. When it says 'external force', the driving force between the wheels and the road, it does just mean the driving force, right? And not the frictional force?
    Also, my calculations for tension have come out at 0N. But I think there must still be a Tension/resultant force between the two vehicles as they are not moving at a constant velocity. Do I need to calculate in car and trailer tension separately?




    2. F=ma
    F-T=ma




    3.
    (a) F=(1000kg+650kg)2.3m/s^2
    =2500N (2sf)
    (b) F-T=ma
    T=F-ma
    T=2500-2500=0
     
    Last edited: Jan 6, 2010
  2. jcsd
  3. Jan 6, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi lemon! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    They're the same …

    by good ol' Newton's third law, action is equal and opposite to reaction, so the (drive) force of the car on the road equals the (friction) force of the road on the car. :wink:
    Yes, you can only find T if it's an external force, so that means that you must consider only the forces on the car (or only the forces on the trailer).

    (So your m in that last equation is wrong.)
     
  4. Jan 6, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi lemon! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    They're the same …

    by good ol' Newton's third law, action is equal and opposite to reaction, so the (drive) force of the car on the road equals the (friction) force of the road on the car. :wink:
    Yes, you can only find T if it's an external force, so that means that you must consider only the forces on the car (or only the forces on the trailer).

    (So your m in that last equation is wrong.)
     
  5. Jan 6, 2010 #4

    tiny-tim

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    Welcome to PF!

    Hi lemon! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    They're the same …

    by good ol' Newton's third law, action is equal and opposite to reaction, so the (drive) force of the car on the road equals the (friction) force of the road on the car. :wink:
    Yes, you can only find T if it's an external force, so that means that you must consider only the forces on the car (or only the forces on the trailer).

    (So your m in that last equation is wrong.)
     
  6. Jan 6, 2010 #5
    Hi:
    So, to answer part b correctly:

    Car: 2500N-T=1000kg x 2.3ms
    T=2500N-2300N=200N

    Trailer: 2500N-T=650kg x 2.3m
    T=2500N-1495N=1005N

    Total T=Tcar+Ttratiler=200N+1005N=1205N
     
  7. Jan 6, 2010 #6
    test
     
  8. Jan 6, 2010 #7
    ms-2
     
  9. Jan 6, 2010 #8

    tiny-tim

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    Hi lemon! :smile:

    I see you had the same problem as I did, with the server!! :biggrin:

    Your car equation is completely correct …

    and you should have stopped there! :wink: :rolleyes:

    Your trailer equation is wrong, because there's only one force on the trailer, and that's T.

    F is a force on the car, not on the trailer … all the trailer gets is what comes through the coupling.

    So you could have said T = mtrailera, which would give you the same result as F - T = mcara … either will do, but the first one is shorter. :wink:
     
  10. Jan 6, 2010 #9
    You rock! Roll! Rave!
    Whatever.
    Thanx
     
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