11.45 Mountaineers often use a rope to lower themselves down the face of a cliff
ID: 1266999 • Letter: 1
Question
11.45 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 75.0-kg climber, who is 1.75m tall and has a center of gravity 1.0m from his feet, rappels down a vertical cliff with his body raised 37.4deg above the horizontal. He holds the rope 1.43m from his feet, and it makes a 20.1deg angle with the cliff face.
A:What tension does his rope need to support?
Express your answer using two significant figures.
B:Find the horizontal component of the force that the cliff face exerts on the climber's feet.
Express your answer using two significant figures.
C:Find the vertical component of the force that the cliff face exerts on the climber's feet.
Express your answer using two significant figures.
D: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?
Express your answer using two significant figures.
Explanation / Answer
you need to solve the following equations
Sum of Fx = 0
Sum of Fy =0
Sum of torques = 0
Suppose the man is the object. He has 3 forces acting on him
Tension, T at A = 20.1 from vertical, gravity straight down (mg), acting at his center of gravity and wall pushing out Fn, and friction on his feet.
Let B = 37.4
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) sin(37.4) + T(1.0) sin(180 - 20.1-37.4) = 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(1.0)sin(180-A-B) = mg(1.0)sin(37.4)
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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