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Homework Help: Inclined Plane and box

  1. May 30, 2010 #1
    The owner of a warehouse ask an engineer to design a ramp which will reduce the force necessary to lift boxes to the top of a 1/2 m step. If there is only room enough for a 4 m ramp, what is the maximum factor by which the lifting force could be reduced?

    I am not sure how to find the maximum factor?? Could anyone help?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 30, 2010 #2


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    hi jerryez! :wink:

    well, the force to lift it without a ramp, vertically upwards (and with zero acceleration), is mgh …

    what is the force needed if you use a 4m ramp? :smile:
  4. May 30, 2010 #3
    Hello Tiny-tim

    Would the force than be equal to mgsintheta?? I dont think this is right but that is my best guess.
  5. May 30, 2010 #4
    You got it...

    Now find theta and compare g to g*sintheta
  6. May 31, 2010 #5
    So theta = 7.2

    g = 9.81 m/s(squared)

    g*sintheta = 1.23

    I still dont understand what the maximum factor by which the lifting force could be reduced is??
  7. May 31, 2010 #6


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    It is a simple machine.

    The mechanical advantage = Load/ effort = Distance traveled by the effort/ Distance traveled by the load. = 4/(1/2)
  8. May 31, 2010 #7


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    hello jerryez! :smile:

    (just got up :zzz: …)

    The reason for the word "maximum" is because it depends on the angle of the force (and also on the acceleration being zero).

    Usually when you push or pull something, your applied force isn't exactly along the slope, and if it's not exact, then you need extra force. :wink:
    hmm … guessing will be no good in the exam :redface:

    the method to use, as nearly always, is good ol' Newton's second law …

    we're assuming no acceleration along the slope

    so (let's call that direction "k") (Ftotal).k = ma.k = 0. :wink:
    Last edited: May 31, 2010
  9. May 31, 2010 #8
    Along with the other things Tim mentioned, maximum would also assume there is no friction. This would make a signficant difference in the real world. So I assume your problem also avoids this variable as well as if you did not pull the mass parallel to the plane or accelerate the mass up the plane.

    I am sorry for the intervention. The posters question went unanswered for a while. So I just added that g v. sintheta*g gives you the appropriate ratio. If its force it would be mg v. m*sintheta*g, but since m is on both sides I took an irresponsible shortcut. Sorry.
  10. May 31, 2010 #9
    I think its says the forced reduced by, so using your shortcut I think it would be 1/2 over 4. If it said how much "easier"...
  11. Jun 5, 2010 #10
    Thank you everyone for your help!! The answer to the problem is 8 I just didnt understand how they got it.

    I also see now how the ratio between g and g*sintheta also results in the correct answer.

    9.81/1.23 = 7.9
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