How Do Climbers Use Physics to Rappel Down Cliffs?

[BlueDevil14]

Homework Statement

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 ). Suppose that an 76.2 kg climber, who is 1.88 m tall and has a center of gravity 1.2 m from his feet, rappels down a vertical cliff with his body raised 33.0 degrees above the horizontal. He holds the rope 1.43 m from his feet, and it makes a 27.2 degree angle with the cliff face.

Homework Equations

The Attempt at a Solution

I understand conceptually how to solve the problem, but I am making a small mistake somewhere that I cannot find.

Using the moment arm method:
Counterclockwise torques: T_y * sin(33)*1.43 m
Clockwise torques: T_x * cos(33)*1.43 m + 746.76 N *sin(33)*1.2

T_x=sin(27.2)*T
T_y=sin(62.8)*T

Because net torque is zero:
sin(62.8)*T*sin(33)*1.43 m = sin(27.2)*T*cos(33)*1.43 m + 746.76 N *sin(33)*1.2

Any help would be greatly appreciated.

 

[mburt]

Are you trying to find the tension in the rope? I'm not clear and what the question is looking for

 

[BlueDevil14]

Yes, I am trying to find tension in the rope. Sorry, I left out that important piece of information.

 

[mburt]

I find the question strange, because really, there is two pivot points. One where the rope is attached to the cliff and also at the feet of the person who is 33 degrees above the horizontal.

How are you deciding whether the force of gravity on the person is clockwise or counterclockwise? And what pivot point are you referring to?

 

[BlueDevil14]

I believe the problem assumes that the rope does not pivot, as odd as that sounds. The torques are all with respect to a pivot about the climber's feet.

Because gravity acts downwards, the torque will be clockwise.

 

[mburt]

You used sin 33 as the angle for force gravity between the person's legs and the horizontal.

This angle is meant to be between the force gravity (which is always straight down) and the person's legs, so instead of using 33, you use 90 - 33 = 57.

Try using:

746.76 N *sin(57)*1.2

For the torque of the person's weight instead

 

[BlueDevil14]

Thanks, that makes sense. For some reason though, it is not correct still.

I assume it is another aimless trig mistake.

 

[mburt]

Sorry I can't be of much more help. I'm only doing entry level physics myself (conveniently, I was doing some torque problems today), but I've never done any torque problems like this.

Truthfully, I'm really confused by these parts:

Because net torque is zero:
sin(62.8)*T*sin(33)*1.43 m = sin(27.2)*T*cos(33)*1.43 m + 746.76 N *sin(33)*1.2

I'm just assuming that's the torque of the rope? But how is it both clockwise and counter clockwise?

And in most likeliness, it's probably a simple mistake

 

[BlueDevil14]

The torque is about the pivot point, I am only using torque as an equation to solve for T