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Mountain Climber Rope Tension

  1. Oct 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist 15 cm to the right of his center of gravity and makes an angle of θ = 23° with the wall. The climber weighs 616 N. The distance from his center of gravity to the wall is 91 cm and the distance from his feet on the wall to the buckle is 106 cm
    (a) Find the tension in the rope
    (b) Find the magnitude and direction of the contact force exerted by the wall on the climber's feet.

    2. Relevant equations
    F = ma
    g = 9.8
    Weight of climber = 616 N = vertical force of ropes tension plus force against feet
    Vertical component of rope tension force = total tension cos 23 degrees
    Force on feet = ?

    3. The attempt at a solution

    616 N = F total * .921 + ?

    I am having real difficulty with this as I don't understand the importance of the center of gravity.
     
    Last edited: Oct 23, 2010
  2. jcsd
  3. Oct 24, 2010 #2

    tiny-tim

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    welcome to pf!

    hi dvolpe! welcome to pf! :wink:
    the importance is that the the weight of the climber acts through the centre of gravity (the centre of mass) …

    when you take moments, the further out the centre of gravity is, the larger the moment (of the weight) will be, and that will change the reaction force at the climber's feet :smile:
     
  4. Oct 24, 2010 #3
    So the sum of the y components of the rope tension, wall force, weight of the climber = 0. The x component of the wall force equals the x component of the tension. Using the buckle as a axis, the sum of the torques equals 0 in the y direction or the y force of wall times .106 m = weight times .091. Force of wall in y direction is 87.17 N. Then substituting in back into first sentence and tension of rope = 572.8 N. The wall force is equal to the square root of the sum of its x and y components. Already have the y component; x component = force of tension in x direction or T cos theta. Solve for wall force. The direction of the wall force is solved by trig: sin theta = y comp/total wall force or theta = 21.2 degrees. Does this approach seem correct? I am concerned about using the correct distances in the sum of torques equation.
     
    Last edited: Oct 24, 2010
  5. Oct 24, 2010 #4

    tiny-tim

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    hi dvolpe! :smile:
    are you treating the x and y directions separately for moments (torques)?

    you can't do that!!

    a vertical force of strength F at perpendicular distance d has the same moment (torque) as a horizontal force of strength F at perpendicular distance d (so long as they're both clockwise or both anti-clockwise)

    more generally, what is important is the direction of the moment … d x F is out of the page, and therefore has to be added to all other moments that are out of the page :smile:

    start again (and, for taking moments, it's easiest if you choose a point which eliminates as many unknowns as possible :wink:)​
     
  6. Oct 24, 2010 #5
    Ok..what I meant to say is that I selected the point of the buckle to take the moment as the forces in the y direction are perpendicular to the point but the forces in the x direction are not and therefore do not contribute to the moment at the buckle. That point gives me the least unknowns. I solved it as I indicated above..is this correct? Thank you.
     
    Last edited: Oct 24, 2010
  7. Oct 24, 2010 #6

    tiny-tim

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    no, at the climber's feet, both the x and y components of the reaction force have a moment about the buckle

    (btw it occurs to me that since there are only three forces on the climber, they must all go through the same point, which instantly gives you the direction of that reaction force :wink:)
     
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