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Falling Mountain Climber

  1. Oct 9, 2006 #1
    Hello,
    I cannot solve this question from 4u physics Dynamic assignment and was wondering if somesome can give me guidance as to how to approach the question?

    84 kg mountain climber rapples down a rope. The rope is damages and can withstand only 675 N of tension. Can the climber limit the descent to a constant speed without breaking the rope? If not, what value can the climber limit the downward acceleration to?

    Thank You!
    Waiting for your reply
     
  2. jcsd
  3. Oct 9, 2006 #2
    You should know that at a constant speed, all forces are in equilibrium. Take it from there.
     
  4. Oct 9, 2006 #3
    Thanks for the reply, but I still don't understand.
    This is how far I got.
    I said FT=675 N
    m=92 kg

    F=ma
    675=92a
    a=7.3 m/s2

    so the in order for the rope to not break the climber has to travel at minimum acceleration of 7.3 m/sec2. In order for him to attain a constant veloctiy he will have to put more tension on the rope and if he does that the rope will break.
     
  5. Oct 9, 2006 #4
    Someone please help!
     
  6. Oct 9, 2006 #5

    radou

    User Avatar
    Homework Helper

    Newton's law of motion states that the resultant (i.e. the sum) of all forces acting on a body equals, in a special case, the product of mass and acceleration, F=ma. Is the force of the rope the only force acting on the climber?
     
  7. Oct 9, 2006 #6
    Well there would be gravity and friction present. So should I add
    g=9.8*92
    =901.6 N
    Fnet=226.6 N (down)
    a=2.5 m/s (down)

    How does that help calculate weather constant velocity is possible?
     
  8. Oct 9, 2006 #7

    radou

    User Avatar
    Homework Helper

    For constant velocity, the net force equals zero, i.e. the forces are in equilibrium. Is this possible?
     
  9. Oct 9, 2006 #8
    Thanks a lot for all that help! This was very helpful for me and saved me a lot of time of trying to understand the question. This is an awesome site with awesome people!
     
  10. Oct 10, 2006 #9
    I had another quick question aboutthe second part: what value can the climber limit the downward acceleration to?

    is what I did before right:
    g=9.8*92
    =901.6 N
    Fnet=226.6 N (down)
    a=2.5 m/s (down)

    or do I have to do the following:
    F=ma
    675=92a
    a=7.3 m/s2

    Can u explain which one and why? I am kind of unsure.
     
  11. Oct 10, 2006 #10
    Can someone please help me again!
     
  12. Oct 11, 2006 #11
    :surprised :

    :confused:
     
  13. Oct 11, 2006 #12
    Remember that it's the total force that gives you the acceleration, not just the force from the rope.
     
  14. Oct 11, 2006 #13
    I Need Major Help Fast, Its About Finding Time!

    How would you get time for this:

    5. A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the ground? (Use GUESS Method)

    G: vi = 8.0 m/s
    g = 9.8 m/s2
    dy = 78.4 m
    U: dx
    E: d = vi*t +1/2 a*t^2
    S: dx = 32 m


    vi =initial velocity
    ^2= squared
    *= times
     
  15. Oct 11, 2006 #14
    srry my bad
     
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