# Lowering a climber with the belay rope routed through a pulley at an angle

• Jpyhsics
In summary, the climber with a mass of 54.4 kg reaches the top of a 15.70 m climbing wall and is lowered down by a belaying person without touching the wall. The rope is inclined at an angle of 44.0° from the vertical and the landing speed should not exceed 2.66 m/s. Using the equations F=ma and v2=v02+2ad, the acceleration is calculated to be 0.2253 m/s2. Considering torque balance on the pulley, the tension force is found to be the same throughout. By assuming that the average force must be constant, the minimum average force exerted by the belaying person is determined to be 521 N.

## Homework Statement

A climber with mass m=54.4 kg has reached the top of the l=15.70 m climbing wall, and then is lowered straight down by the belaying person without touching the wall. Assume that at the top the rope is going through the pulley, while to the belaying person the rope is inclined by the angle of θ=44.0° from the vertical. If the landing speed of the climber should not exceed 2.66 m/s, what is the magnitude of the minimum average force that is exerted by the belaying person? Express your answer in Newtons.

1. F=ma

## The Attempt at a Solution

•found the acceleration to be 0.2253..m/s2 using equation 2
•Have another equation for Fnet=FT-mg
unsure where the angle comes into play, and whether finding FT is the force exerted by the belayer.

Jpyhsics said:
whether finding FT is the force exerted by the belayer.
What do you think? Can you work it out by considering torque balance on the pulley?

haruspex said:
What do you think? Can you work it out by considering torque balance on the pulley?
Well I thought the acceleration was the same throughout the pulley so does that mean that the force is the tension force is the same?

Jpyhsics said:
Well I thought the acceleration was the same throughout the pulley so does that mean that the force is the tension force is the same?
The tension can be taken as the same, but not for that reason.
If the tensions are different then the torques they exert on the pulley will not cancel. If the pulley is taken as having no moment of inertia and no axial friction then that would lead to an infinite angular acceleration.

So what answer do you get?

There is a subtlety with this question which I suspect the question setter has overlooked. To answer it rigorously you would need to prove that for the average force to be minimised it must be a constant force. For now, just assume that it is.

haruspex said:
The tension can be taken as the same, but not for that reason.
If the tensions are different then the torques they exert on the pulley will not cancel. If the pulley is taken as having no moment of inertia and no axial friction then that would lead to an infinite angular acceleration.

So what answer do you get?

There is a subtlety with this question which I suspect the question setter has overlooked. To answer it rigorously you would need to prove that for the average force to be minimised it must be a constant force. For now, just assume that it is.
I have found my answer to be 546 N. Is that correct?

Jpyhsics said:
I have found my answer to be 546 N. Is that correct?

Jpyhsics said:
Jpyhsics said:
Fnet=FT-mg
I say probably because it depends what sign convention you are using. Are you defining Fnet to be a positive value or are you defining it to be measured in the upwards direction (so negative if actually downwards).

Jpyhsics said:
Hi Jphysics,

Please type out your work equations so that helpers can directly quote them, or at the very least make your images local to PF (UPLOAD button) and viewable in-thread. Thanks.

haruspex said:

I say probably because it depends what sign convention you are using. Are you defining Fnet to be a positive value or are you defining it to be measured in the upwards direction (so negative if actually downwards).
I'm saying negative is downwards. Also, would I need to use the angle at all?

Jpyhsics said:
I'm saying negative is downwards.
So what sign should Δx have?
Jpyhsics said:
would I need to use the angle at all?
No.

haruspex said:
So what sign should Δx have?

No.
oh so would you think my revised answer of 521N be right?

Jpyhsics said:
oh so would you think my revised answer of 521N be right?
Yes.

## 1. How does routing the belay rope through a pulley affect the lowering process?

Routing the belay rope through a pulley at an angle can make the lowering process smoother and more controlled. The pulley reduces friction on the rope, making it easier for the belayer to control the speed of the descent.

## 2. Is it safe to lower a climber with the belay rope routed through a pulley?

Yes, it is safe to lower a climber with the belay rope routed through a pulley. In fact, using a pulley can make the lowering process safer by reducing the risk of the rope slipping or the belayer losing control.

## 3. How does the angle of the pulley affect the lowering process?

The angle of the pulley can affect the amount of friction on the rope and the speed of the descent. The more acute the angle, the less friction and slower the descent. A larger angle can increase friction and result in a faster descent.

## 4. Are there any special techniques or precautions to take when lowering a climber with the belay rope routed through a pulley?

Yes, there are a few things to keep in mind when lowering a climber with a pulley. First, make sure the pulley is properly attached and aligned with the direction of the rope. Second, communicate with the climber to ensure a smooth descent. And third, be prepared to adjust the angle of the pulley as needed to control the speed of the descent.

## 5. Can a pulley be used for lowering on any type of climbing route?

Yes, a pulley can be used for lowering on any type of climbing route, including sport, trad, and top-rope routes. However, it is important to always assess the specific situation and adjust the angle of the pulley accordingly to ensure a safe and controlled descent.