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Heavy Load problem

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A truck with a heavy load has a total mass of 5100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2.

    What was the mass of the load? Ignore rolling friction.
    Express your answer with the appropriate units.

    3. The attempt at a solution
    Force parallel = 5100 * 9.8 * sin 15 = 12935.78 N
    Let m be mass of load. The final mass of the truck is 5100 – m.
    Since the truck is accelerating at 1.5 m/s^2, let’s multiply this by 1.5.

    F = (5100 – m) * 1.5 = 7650 – 1.5 * m
    Set this equal to 12935.78 and solve for m.
    7650 – 1.5 * m = 12935.78
    1.5 * m = 5285.78
    m = 5285.78 ÷ 1.5
    This is approximately 3523.853 kg.
  2. jcsd
  3. Oct 15, 2016 #2
    Why is this wrong ?
  4. Oct 15, 2016 #3


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    Staff: Mentor

    I think there is a term missing. Think about the situation. If the mass of the load M_L were zero, you should get zero acceleration for the equation relating net acceleration to M_L. If you write the equation for acceleration a as a function of M_L given the way you have laid it out so far, you would get something like this:

    [tex]a = \frac{F_0}{5100 - M_L}[/tex]

    You can see that as M_L goes to zero in this equation, you still have an acceleration. What term might be missing from this equation that would make a-->0 as M_L -->0? Does that help you re-write your equations?
  5. Oct 15, 2016 #4
    F = (5100 – m) * 1.5 = 7650 – 1.5 * m so here is were I messed up ?
  6. Oct 15, 2016 #5


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    Staff: Mentor

    I think it might help to write the two sets of equations, for before and after the load has fallen off. Write out

    [tex]∑F = ma[/tex]

    for before and after. Be sure to include all forces acting on the truck... :smile:
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