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Truck and box

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    a truck of mass 2000 kg traveling 20 m/s to the right carries a load of mass 500 kg positioned 3 meters behind the driver. The load is kept on the truck by friction, and the coefficient of friction between the load and the bed of the truck is 0.5. What is the shortest distance in which the truck can stop without having the load slide forward enough to hit the driver? (Note that the load can slide forward 3 meters before it hits the driver.)


    2. Relevant equations



    3. The attempt at a solution

    to solve this question,

    take truck moving right = +ve

    the change in momentum of the box = 0 - (500)(20) = -10000 kgms-1

    hence, the Force (Fc) that brings about this change = -10000/t

    assuming the box doesn't move after the truck stops, hence Fc = Friction force of box on truck
    <or should it be Forward momentum force acting right = friction force acting left on box?>
    <because my Fc = -ve, hence it tells me its acting left. so it can't be = friction force acting left, to cancel out each other so that box doesn't move>
    since friction acts in the direction of the aceleration, and the box is acelerating to the left as the truck slows down.

    anyway, therefore Friction = (500)(9.8)(0.5) = 2450N
    thus t = 10000/2450 = 4.1s
    assume truck decelerates constantly, 0 - 20/ t = a , => a = -20/4.1 = -4.9ms-2
    finally, use s=ut+1/2at^2 , => s = 20(4.1)+1/2(-4.9)(4.1^2) = 40.8m

    however, the question allows the box to move 3m and i realise that the final answer of 37.8m is just 40.8- 3 = 37.8m.

    so is this a coincidence? or is there some reason behind it? because i still don't know how to factor in the 3m allowance

    from my understanding, the 3m allowance would come from the net force (Fnet) between (Fc) and Friction, as (Fc) will be larger than Friction now for the box to move 3m. but i don't know how to relate Force to distance moved.

    i tried using 1/2mv^2 of the box = work done against friction + work done to move box 3m. but it doesn't work.

    thanks for the help!
     
  2. jcsd
  3. Sep 24, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi quietrain! Welcome to PF! :smile:
    Yup! big coincidence! :biggrin:

    Start again …

    if the acceleration of the truck is -a, what is the acceleration of the load? and how far does each go before it stops? :smile:
     
  4. Sep 24, 2009 #3
    erm, if the truck has -a, then the box would have -a too.

    the distance moved by the truck = distance moved by the box = s = ut + 1/2at^2
    s= 20(t) + 1/2(-a)t^2

    so i don't have t and don't have -a.
     
  5. Sep 24, 2009 #4
    oh, am i right to say that the only force that causes the change in momentum of the box from 10000kgms-1 to 0kgms-1 is friction? assuming no air resistance.

    therefore 10000/t = friction= 2450N

    time taken for the box for this change of momentum is t = 4.1s regardless of how the truck slows down ?

    so deceleration of box due to friction =>a = 0 -20 /4.1s = -4.9ms-2

    so distance moved by box = s = (20)(4.1s) + 1/2(-4.9)(4.1^2) = 40.8

    so the box would move a distance of 40.8m in 4.1s regardless of how the truck moves?

    so the truck needs to move at least 37.8m to prevent the box(after it moves 3m) from hitting the driver ?
     
  6. Sep 25, 2009 #5

    tiny-tim

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    Hi quietrain! :smile:
    No, if the box is slipping relative the truck, then it has a different velocity, and a different acceleration, doesn't it? :wink:

    Find this acceleration using the friction force, and good ol' Newton's second law :smile:
     
  7. Sep 25, 2009 #6
    so am i right to say that the only force that causes the change in momentum of the box from 10000kgms-1 to 0kgms-1 is friction? assuming no air resistance.

    therefore 10000/t = friction= 2450N

    time taken for the box for this change of momentum is t = 4.1s regardless of how the truck slows down ?

    so deceleration of box due to friction =>a = 0 -20 /4.1s = -4.9ms-2

    so distance moved by box = s = (20)(4.1s) + 1/2(-4.9)(4.1^2) = 40.8

    so the box would move a distance of 40.8m in 4.1s regardless of how the truck moves?

    so the truck needs to move at least 37.8m to prevent the box(after it moves 3m) from hitting the driver ?
     
  8. Sep 25, 2009 #7

    Doc Al

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    Staff: Mentor

    That's right.
     
  9. Sep 25, 2009 #8
    thanks!
     
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