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Homework Help: Would like to double check my work. Mostly F=ma if I'm right.

  1. Aug 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A) A car of mass 1000 kg is acted on by a net force of 2500 N. What is the resulting acceleration?

    B) A car of mass 1000 kg pulls a caravan of mass 800 kg; the driving wheels of the car exert a force of 8000 N on the road. The total resistance to motion is 3000 N.
    B1) What is the net accelerating force?
    B2) What is the acceleration?
    B3) What is the force of the car on the caravan?

    C) An engine of mass 5000 kg pulls a train of ten trucks each of mass 2000 kg along a horizontal track. Assume that the frictional forces to be 5000 N and that the engine exterts a force of 50,000 N on the rails. If the trucks are numbered from 1 to 10 starting with the one next to the engine calculate:
    C1) The net total acceleration force
    C2) The acceleration of the train
    C3) The force of truck 6 on truck 7
    C4) The force of truck 9 on truck 8

    D) A rescue helicopter lifts a strecther case off a hillside. The injured climber and stretcher have a mass of 180 kg and the rope supporting them has a breaking force of 2000 N. Calculate:
    D1) The tension in the rope when the stretcher and climber are suspended from the rope at rest
    D2) The maximum vertical acceleration possible before the rope breaks

    2. Relevant equations
    All I know and all that we have learned so far equation-wise is that F= ma and the pendulum equation.

    3. The attempt at a solution

    A) F = ma
    2500 N = 1000 kg * a
    2500/1000 = 1000a/1000
    2.5 m/s^2 = A

    m1 = 1000 kg
    m2 = 800 kg
    F = 8000 N
    resistance to motion = 3000 N

    F= 5000N

    B2) F = ma
    5000 = 1800 * a
    2.78 m/s^2

    F = 800 (2.78)
    F = 2224 N

    C) I am pretty sure this is all wrong, but here's my attempt.
    Engine = 5000 kg
    Trucks = 2000 kg ( * 10)
    frictional F = 5000 N
    engine F = 50,000 N

    C1)F = 45,000 N

    C2) F = ma
    45,000 = 20,000 * a
    a = 2.25 m/s^2



    m = 180 kg
    F = 2000 N

    D1) F = ma
    F= 180 (9.8)
    F = 1764 N

    D2) F = ma
    2000 = 180 *a
    11.11 m/s^2 = a

    Thanks in advance!
    Last edited: Aug 21, 2009
  2. jcsd
  3. Aug 21, 2009 #2
    Seems right to me!!!
  4. Aug 21, 2009 #3


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    (A) looks good.

    If I told you that the mass is not 1800 kg, can you figure out why?

    For (C1), they are asking for a force here, not the acceleration.
  5. Aug 21, 2009 #4
    Thank you both. Yeah I made the same mistake on B earlier, then I edited it. I wasn't reading carefully.

    Redbelly98, are you saying that B3 is wrong? Or simply asking if I can figure it out without the given 1800 kg?
  6. Aug 21, 2009 #5


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    B3 is wrong. They ask for the force on the caravan, so use the mass of the caravan. (Hint: it's not 1800 kg)
  7. Aug 21, 2009 #6
    I get that you summed the forces along the x-axis to be positive 50 000N in the positive direction and the friction as -5000N, so i'm no expert but that should be right. However shouldn't the engines weight be added to the mass (20,000+5000 kg) * a ?

    I reckon this has to be wrong although i'm not dead sure how to do it right. I am guessing that if you assume the friction is evenly distributed along the rail then the further back you go the less force will need to be exerted between different trucks.
    Last edited: Aug 21, 2009
  8. Aug 21, 2009 #7
    So since the caravan's mass is 800 kg,

    F = 800 (2.78)
    F = 2224N
  9. Aug 21, 2009 #8


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    Yup, looks good :smile:
  10. Aug 21, 2009 #9

    I'm also trying to work out C3 and C4. I already turned in my homework earlier this morning, but I am going to try to remember how I changed it.

    I remember things were different between each car; I believe I got -4000N from truck 6 to 7 and 4000N from 9 to 8. Something like that.

    I'm working it out right now...


    Okay now I am confusing myself. Was I supposed to take into account the engine weight?
  11. Aug 22, 2009 #10
    I think so.. but i'm not a pro or anything, just a lowly student.
  12. Aug 22, 2009 #11


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    Yes. Also, this problem hinges on drawing your free-body diagrams wisely. For example, if you're asked for the force of 6 on 7 and you draw a free-body diagram for 6 or 7, you'll have to include the contact force with 5 or 8, and that's not any easier to find. If you draw a FBD for the section of the train after carriage 6, you'll have only one contact force: the one you're trying to find.

    Lastly, you have to decide how to calculate friction. Is the 5000 N evenly distributed between the carriages? Does the engine get more because it's heavier and is fighting air resistance? The question doesn't say, so you'll have to choose what to assume.
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