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Rotational equilibrium question

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff. Suppose that an 82.0 kg climber who is 1.90m tall and has a center of gravity 1.1 m from his feet rappels down a vertical cliff with his body raised 35.0° above the horizontal. He holds the rope 1.40m from his feet, and it makes a 25.0° angle with the cliff face.

    (a) What tension does his rope need to support?
    (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet.
    (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?


    2. Relevant equations
    [itex]\sum\tau=0[/itex]

    [itex]\sum F=0[/itex]

    [itex]\tau=rxF = rFsin \theta[/itex]

    [itex]f_k=\mu_kn[/itex]


    3. The attempt at a solution
    My teacher went through the problem with us, but I only have his notes as I was sick that day, and I don't quite understand everything. The following is his work:

    (a) [itex]\sum\tau=0[/itex]

    [itex]T(1.4m)cos(10°) - mg(1.1)cos(35°)=0[/itex]

    [itex]T=525 N[/itex]

    I'm confused as to how exactly he figured out to use the 10° with cos? I understand that the angle between the rope and the person is 100° and in order to calculate torque, we must use perpendicular (90°) forces, so 100°-90°=10°, but how did he know to cos it?

    (b) and (c) I understand. Thanks!
     
  2. jcsd
  3. Dec 11, 2011 #2
    Right, the rope makes an angle of 100° with the body. With this in mind, you can draw a component vector that begins from the body and goes up to the tip of the tension vector, in such way that the component is perpendicular to the body.
    You have then a 90°-80°-10° triangle, from which you can find the component of tension for torque: T cos 10°. You could also use T sin 80°. :)
     
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