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Newton's Third Law: A Car Pushes A Truck

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A 1000 kg car pushes a 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N. Rolling friction can be neglected.


    2. Relevant equations

    F=ma

    acceleration of car = acceleration of truck


    3. The attempt at a solution

    I have determined that both the car and the truck must move at the same acceleration. The car pushes off the ground with 4500N, so that:

    F = ma
    4500 = (1000)a
    a = 4.5 m/s^2


    I attempted to solve for F on the truck:

    F = ma
    F = (2000)(4.5)
    F = 9000

    But this is incorrect. Should I be combining the masses so that F = (3000)(4.5)? Or am I just way off here?

    I am struggling to put together the proper free-body diagram and I think its messing up my understanding of what force is on what object. A push in the right direction is appreciated! Pun intended.
     
    Last edited: Oct 15, 2007
  2. jcsd
  3. Oct 15, 2007 #2

    PhanthomJay

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    You might want to draw a FBD of the enture truck-car system. What is the net force acting on the system? What is the mass of the system? Your FBD's are not correct since you have not properly identified all forces acting individually on the isolated FBD of the car or truck.
     
  4. Oct 15, 2007 #3
    Hmmm, not exactly sure what you mean by a FBD of the entire system. I assume it would be a particle with a normal force, a weight force (consisting of the 1000 kg car + the 2000 kg truck), and the force of the car's tires on the ground (pointing to the right). That should be it since friction is neglected.

    I haven't seen anything that combines the system into one FBD in my book... not sure how it helps me isolate the force on the truck?
     
  5. Oct 15, 2007 #4

    PhanthomJay

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    It appears that you have correctly identified the forces acting on the car-truck system. The only unbalanced force acting on the system is the 4500N force acting to the right. So what's stopping you from using Newton 2 for the system? That will give you the acceleration, then you can look at the truck by itself to find the net force acting on it. NOTE: You draw an FBD of the system by drawing a circle around the car and truck and identifying the forces acting external to it.
     
  6. Oct 15, 2007 #5
    Ok, I think I figured it out:

    Fnet Car:

    4500 - Ftruck = (mass of car)(acceleration)

    Fnet Truck:

    Fcar = (mass of truck)(acceleration)

    Solve both for acceleration and set them equal to each other, remembering that Fcar = Ftruck from Newton's 3rd Law.



    (4500-Ftruck)/(mass of car) = (Fcar)/(mass of truck)

    Ftruck = Fcar = 3000N
     
  7. Oct 15, 2007 #6

    PhanthomJay

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    Yes, that will do it. As a double check, look at the system, Fnet = (m + M)a ; 4500 = 3000a, solve a =1.5 and then Fcar on truck = 2000(1.5) = 3000.
     
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