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Homework Help: Boulder being pulled up by a chain. Find max acceleration

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A 800 kg boulder is raised from a quarry 150 m deep by a long uniform chain having a mass of 580 kg. This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.90 times its weight without breaking.

    What is the maximum acceleration the boulder can have and still get out of the quarry?

    How long does it take to be lifted out at maximum acceleration if it started from rest?


    How do I find time?

    2. Relevant equations

    f=ma, one of the kinematic equations for the second question

    3. The attempt at a solution
    The max force is is the weight of the chain*2.9 = 16483.6

    The weight of the system is (chain*9.8)+(boulder*9.8) = 13524

    I'm not sure where to go with these numbers. The weight of the system would be pointing down. So something with the tension would be the force going up. How do I find the tension, and then move to acceleration?

    EDIT EDIT EDIT: The tension to hold up the boulder = the weight of the boulder and chain. So I did (Max tension - tension value)=18483.6 - 13524=2959.6

    2959.6 is the force you can have for the system to work. So 2959.6/(mass of the boulder and chain) = 2959.6/1380 = 2.14

    Is 2.14 right?
    Last edited: Feb 9, 2009
  2. jcsd
  3. Feb 9, 2009 #2
    In one place you write that the maximum force is 16483.6 (no units given so I suppose we are in penguins/cubic banana as usual).
    Later on we use the value 18483.6. So a little discrepancy has crept in.
  4. Feb 19, 2009 #3
    You can use the equation X=X0 + V0*t + 1/2*a*t2
    Since you know the initial distance and initial velocity are 0, you solve for just X=1/2*a*t2

    You know X and a. Just rearrange and solve for t.
  5. Feb 19, 2009 #4
    delta X = Vot + 1/2 at^2. Vo is 0.
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