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Find mass of load that falls off truck

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  1. Mar 2, 2017 #1
    1. The problem statement, all variables and given/known data
    A truck with a heavy load has a total mass of 7500 kg. It is climbing a 15 degree 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/s 2 . What was the mass of the load? Ignore rolling friction.

    2. Relevant equations
    ΣF = Fg + n + F1 = ma , where n is the normal force and F1 is the parallel force

    3. The attempt at a solution
    Before the load falls off we know a = 0 therefore
    ΣF = Fg + n + F1 = 0

    Fg before fall = 9.81 m/s^2 * 7500 kg = 73575 N

    Fgx before fall = -73575*Sin(15) = -19042

    Because ΣFx = 0 before fall, Fgx = -F1x

    => F1x = 19042 N

    After fall we have:

    Fx = 19042 + Fgx = FgSin(15) = m*1.5
    Fy = 0

    I'm not sure where I need to go from here. I need to find m so I can find the difference, but I can't do this without either the magnitude of Fg or Fgx. Not sure how I can find these...
     
  2. jcsd
  3. Mar 2, 2017 #2

    TSny

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    Welcome to PF!

    Is there a way to express Fg in terms of m?
     
  4. Mar 2, 2017 #3
    Ok so I assume you mean Fg = ma = 9.18m

    I guess this gives me these simultaneous equations:

    Fg = 9.81m
    Fg = 1.5m / Sin(15)

    But the solution of this is Fg = m = 0, so there's something wrong o_O
     
  5. Mar 2, 2017 #4

    TSny

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    OK

    How did you arrive at this?
     
  6. Mar 2, 2017 #5
    Fx = 19042 + Fgx = FgSin(15)
    Fy = 0
    => FgSin(15) = Fx = F = ma = 1.5m
    => Fg = 1.5m/Sin(15)
     
  7. Mar 2, 2017 #6

    TSny

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    The first equality looks good. But I don't understand the second equality. How did you get 19042 + Fgx = FgSin(15)?
     
  8. Mar 2, 2017 #7
    My understanding was that Fx = Fgx = FgSin(15)
     
  9. Mar 2, 2017 #8
    Oh, wait, I see why that is wrong haha
     
  10. Mar 2, 2017 #9
    Fx = 19042 + Fgx = 19042 + FgSin(15) = 1.5m

    which gives the simultaneous equations:

    Fg = 9.81m
    Fg = (1.5m - 19042)/ Sin(15)

    With solutions:

    f = -46249 and m = 4714.52

    Therefore:
    Δm = 7500 - 4714.52 = 2785.48 kg

    Yes?
     
  11. Mar 2, 2017 #10

    TSny

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    Does Fgx = +FgSin(15) or does Fgx = - FgSin(15)?


    Basically this is correct except for sign errors stemming from the wrong sign for Fgx. I think your results for m and Δm are correct.
     
  12. Mar 2, 2017 #11
    Ah yes, you are indeed right. Thankyou for your help :)
     
  13. Mar 2, 2017 #12

    TSny

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    OK. Good work.

    When solving the two equations

    Fg = 9.81m
    19042 - FgSin(15) = 1.5m

    it might be easier to just substitute the expression for Fg from the first equation into the second equation. So, the second equation becomes

    19042 - 9.81m Sin(15) = 1.5m

    Then you can solve this for m. But your method of doing the algebra is OK.
     
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