Mountaineers often use a rope to lower themselves down the face of a cliff (this

Question

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 (the figure (Figure 1) ). Suppose that an 76.7-kg climber, who is 1.70m tall and has a center of gravity 1.4m from his feet, rappels down a vertical cliff with his body raised 38.2? above the horizontal. He holds the rope 1.55m from his feet, and it makes a 29.6? angle with the cliff face.

AWhat tension does his rope need to support?

BFind the horizontal component of the force that the cliff face exerts on the climber's feet.

CFind the vertical component of the force that the cliff face exerts on the climber's feet.

DWhat 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?

Explanation / Answer

you need to solve the following equations
Sum of Fx = 0
Sum ofFy=0
Sum of torques = 0

Suppose the man is the object. He has 3 forces acting on him
Tension, T at A = 24.2 from vertical, gravity straight down (mg), acting at his center of gravity and wall pushing outFn, and friction on his feet.
Let B = 28.9
Lets look at the x direction:
right is positive
Sum Fx = -Tsin(A) + Fn = 0 (draw triangles to see why it is sin)
Now in the y direction, let up be positive
Sum Fy = Tcos(A) - mg +Ffr= 0
note friction acts up, since his feet want to slip down....

Note, at this point we have 3 unknowns (what are they?) so we need another equation. We determine the torques acting about the the man's feet.

We choose his feet as the axis of rotation, since Fn and Friction will not contribute torques about this point (why not?)

Using the equation for torque = distance x force x sin(Angle btween d and F)

We will assume that both the tension and the mass act at the center of gravity, or 1.3 m from the feet. So we see
take torques that tend to rotate counter clockwise as positive

torque = - mg(1.3) sin(28.9) + T(1.3) sin(180 - 24.2-28.9) = 0
You may have to think about that last angle. You need to find the angle between the line that is the rope and the line that is his legs. recall 180 deg in a triangle. So to sum up

T(1.3)sin(180-A-B) = mg(1.3)sin(28.9)
Tcos(A) - mg + Ffr= 0
-Tsin(A) + Fn = 0

Note to answer you just need first eqn
to answer b you need Fn (use 3rd eqn and T from a)
to answer c you need to find Ffr
finally to answer d) recall Fr = uFn (from b)

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