Forces on abseiling (rappelling) rock climber: find vertical reaction force

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SUMMARY

The forum discussion centers on calculating the vertical reaction force experienced by a rock climber during abseiling, specifically using the weight of the climber and the tension in the rope. The original poster (OP) attempted to solve the problem using the equation Cos(70°) x 590 N, yielding a reaction force of approximately 202 N. Participants emphasized the importance of balancing forces, including weight and tension, and clarified that the reaction force is not solely equal to the weight of the climber but must consider all acting forces. The discussion highlights the need for clear conventions in force direction when solving mechanics problems.

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This discussion is beneficial for A-level physics students, educators teaching mechanics, and anyone interested in understanding the forces acting on climbers during abseiling.

Viraj Vasantla
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Homework Statement


Can someone please help me with this A level physics question on Mechanics.
Help will be appreciated. The picture of question is attached. :D
Thank you

Homework Equations

The Attempt at a Solution


[/B]sin20x590=201.79N or cos70x590=201.79N
 

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Welcome to PF. In order for us to help, you need to make a serious attempt to solve the problem yourself. How far can you get yourself?
 
Yes i have made a serious attempt. In my attempt, i saw that the reaction force is equal to the weight of the person. Therefore the component of weight should be equal to vertical component of the reaction force. So, the answers came to Cos(70)x590 which equals 202N to 3 S.F but i believe i may be incorrect so i posted this thread to ask for help and see what the correct answers is.I would appreciate it if someone checked if i am correct or not.
 
Viraj Vasantla said:
Yes i have made a serious attempt. In my attempt, i saw that the reaction force is equal to the weight of the person. Therefore the component of weight should be equal to vertical component of the reaction force. So, the answers came to Cos(70)x590 which equals 202N to 3 S.F but i believe i may be incorrect so i posted this thread to ask for help and see what the correct answers is.I would appreciate it if someone checked if i am correct or not.

Why would the reaction force be equal to the weight of the climber?
 
I'm not sure but doesn't the vertical vector Fr be equal to all the other vertical vector(in the opposite direction off course)
Freactiony= Fmass + Ftensiony
 
What past paper or question bank is this from please
 
Mathijsgri said:
I'm not sure but doesn't the vertical vector Fr be equal to all the other vertical vector(in the opposite direction off course)
Freactiony= Fmass + Ftensiony
Yes, those three forces must balance, but perhaps not in quite that relationship.
You need to specify the convention you are using for force directions. Are you taking all forces as positive upward (so a force that turns out to be downward will be negative) or are you taking each force as positive in the direction in which you expect it to act (e.g. down for gravity, up for tension)?

But anyway this is different from your earlier statement that the reaction force equals the weight of the person. At least now you have included tension.
 
donnyboy9999 said:
What past paper or question bank is this from please
Well, this thread is from a year and a half ago, and the original poster has not been back since then. So you probably will not get an answer to your question..

From the OP's Profile Page:
Viraj Vasantla was last seen:
Sep 28, 2016
 
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