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Kinetics/kinematics question

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    a 3.5 Mg engine is suspended from a spreader beam having a negligible mass and is hoisted by a crane which exerts a force of 40kN on the hoisting cable. determine the distance the engine is hoisted in 4 s starting from rest.

    2. Relevant equations

    f=ma
    vf^2 = vi^2 + 2ad
    v=Vo + at


    3. The attempt at a solution

    first i used f=ma
    converted force of 40kN to N so 40000 and converted 3.5 Mg to kg so 3500
    then a = f/m so a = 11.4286 m/s^2 (this seems to high but i continued anyways)

    then i took v= Vo +at Vo=0 a is from above and t = 4s just plug in and you get 45.71 m/s

    then use
    vf^2 = vi^2 + 2ad vf = 45.71 vi = 0 a = 11.43 solve for d and you get something like 91m that i know is wrong. i should get something like 13 m does anyone see where im going wrong??
     
  2. jcsd
  3. Nov 15, 2009 #2
    There is more than one force acting on the engine. You are only accounting for the force of the crane that is why the net force is high.
     
  4. Nov 15, 2009 #3
    oh ok... i got it now i forgot to take away the force of the crate pulling down on the crane. when i do that i get the right answer now.
    thanks alot srmeier!!!!!!
     
  5. Nov 15, 2009 #4
    My pleasure ^^

    Side note: (One force is caused by the collection of hoisting cables which are connected to the engine & the other is the force of gravity on the engine. also note that the tension force is only in the y-direction because the x-components of force cancel one another.)
     
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