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Obstacle course acceleration problem

  1. Jan 11, 2010 #1
    1. The problem statement, all variables and given/known data
    While running an obstacle course, Jane comes to a rope hanging from a platform. She grabs a rope, which then accelerates her up into the air. Her mass is 55 kg and she accelerates upward at 1.50 m/s2. The tension on the rope is ____N.


    2. Relevant equations
    F=ma


    3. The attempt at a solution
    F=55(1.50)
    F=82.5 N

    Is this right?
     
  2. jcsd
  3. Jan 11, 2010 #2
    Re: Acceleration

    No, when applying Newtons 2nd law you need to consider the sum of the forces acting on a body.

    So,

    [tex]\sum F = ma[/tex]
     
  4. Jan 11, 2010 #3
    Re: Acceleration

    can you show me how to do a question like this?
    (keep in mind i only have physics 20-1)
     
  5. Jan 11, 2010 #4
    Re: Acceleration

    What forces are acting on the object when she grabs the rope?

    Let me remind that we are given her acceleration in the upwards direction so we really only need to consider the vertical forces acting on the body.
     
  6. Jan 11, 2010 #5
    Re: Acceleration

    the force of her and the force of gravity?
     
  7. Jan 11, 2010 #6
    Re: Acceleration

    The force of tension and the force of gravity. The force of gravity can also be denoted as her weight.

    Now apply newtons 2nd law and solve for the force of tension.
     
  8. Jan 11, 2010 #7
    Re: Acceleration

    F=539.55N
    is that the ans?
     
  9. Jan 11, 2010 #8
    Re: Acceleration

    No.

    How about instead of posting one line of "Is this the answer?" you show us your work in full detail and your reasoning behind it.

    Only then will we be able to truly help you.

    (By the way, I don't consider handing you the answer helping)
     
  10. Jan 11, 2010 #9
    Re: Acceleration

    dude i did what u told me to do i wasnt randomly posting that
    but if u cant see that..

    F=ma
    F=her mass (aka 55kg, since you may not wan to scroll to the top)(9.81m/s2)
    F=539.55 N

    (yes im grouchy but its only cus ive only been through this with other ppl who after a waste of my time can't figure it out themselves in the end, i apologize)
     
  11. Jan 11, 2010 #10
    Re: Acceleration

    [tex]\sum F = ma[/tex]

    It's the SUM OF THE FORCES on the body that will correspond to its mass times its acceleration.

    You told me there were two forces acting on her body (atleast in the vertical direction).

    If you defined the y axis to be positive upwards and negative downwards, sum the forces on the left handside of the equation according to their direction and on the right hand side simply compute her mass times her given acceleration.(keeping in mind the way you've defined your y axis)
     
  12. Jan 12, 2010 #11
    Re: Acceleration

    so

    (1.50)(55)-(9.81)(55)=-524.835N
     
  13. Jan 12, 2010 #12
    Re: Acceleration

    [tex]T - F_{g} = ma[/tex],

    You need to review/practice your application of Newton's 2nd law. Are you drawing free body diagrams when you're attempting a problem?
     
  14. Jan 12, 2010 #13
    Re: Acceleration

    So T=622.05

    i didnt get that from no where i went T=ma+Fg
     
  15. Jan 12, 2010 #14
    Re: Acceleration

    [tex]F_{g} = mg[/tex]
     
  16. Jan 12, 2010 #15
    Re: Acceleration

    so T=(55)(1.50)+(55)(9.81)
     
  17. Jan 12, 2010 #16
    Re: Acceleration

    Correct.
     
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