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

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Jpyhsics
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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.

Homework Equations


1. F=ma
2. v2=v02+2ad

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.
 
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Discussion

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?
Seems too much. Please post your working.
 
Jpyhsics said:
Here is a link: https://drive.google.com/open?id=131I9txLIKVypnj2XrLR93tB-Zalq2WLR
Your mistake is probably here:
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).
 
haruspex said:
Your mistake is probably here:

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?