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Finding force from position equation

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A 4.50kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t)is applied to the end of the rope, and the height of the crate above its initial position is given by ##y(t) = (2.80m/s)t +(0.61m/s^3)t^3## What is the force at 4 seconds?

    2. Relevant equations
    $$F = ma$$

    3. The attempt at a solution
    Taking the double derivative to get an acceleration equation:
    ##y'(t) = (3*0.61) t^2 + 2.80##
    ##y''(t) = a(t) = (0.61*3*2) t = 3.66t##

    ## F = 4.50 kg * 3.66 * 4.00s = 65.9 N##

    Although MasteringPhysics tells me my answer's wrong, it doesn't give any explanation.
     
    Last edited: Mar 8, 2015
  2. jcsd
  3. Mar 8, 2015 #2

    rude man

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    Homework Helper
    Gold Member

    Your last equation says force = mass x acceleration x time. Is that what newton said?
     
  4. Mar 8, 2015 #3
    I thought 3.66t would give the acceleration since the original equation gives the jerk.
     
  5. Mar 8, 2015 #4

    gneill

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    Staff: Mentor

    One issue is that you didn't carry the units of the constants through your differentiations. Thus your constant in the last equation should be 3.66 m/s3.

    A second issue is that I don't see where the force due to gravity on the crate is taken into account.
     
  6. Mar 8, 2015 #5
    Ah! Thank you very much. That was the missing piece.
     
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