Critical angle- two climbers in a roped party

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Homework Help Overview

The problem involves two climbers on a frosted hillside, focusing on the critical angle of inclination to prevent slipping. The scenario includes different conditions based on which climber is slipping and the coefficients of friction between their shoes and the ice.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law for both climbers and the role of static and kinetic friction in determining the critical angle. Questions arise regarding the correctness of calculated angles and the approach to take when switching the roles of the climbers.

Discussion Status

Participants are actively engaging with the problem, verifying calculations, and exploring different interpretations of the scenario. Some guidance has been offered regarding the application of forces and the use of equations, though no consensus has been reached on the final angles.

Contextual Notes

Participants are working under the constraints of given coefficients of friction and the masses of the climbers, while also considering the implications of switching roles in the problem setup.

mmoadi
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Homework Statement



-Two climbers are climbing up the frosted hillside in a roped party. How much does the most inclination have to be for them not to slip?
-Second climber starts to slip downwards because he was not careful. The first climber notices this at once and tries to restrain the fall with the rope that connects them. How much can the inclination of the slope be the most for the first climber not slip too?
-How much does the angle have to be if the roles were switched, the first climber slipping, and the second climber is trying to restrain him?
-Coefficient of kinetic friction between the shoes and ice is 0.1, and the coefficient of static friction is 0.2. The mass of the first climber is 100 kg and the mass of the second climber is 50 kg. Treat both climbers as wooden blocks.

Homework Equations



tanθ= µ_s

The Attempt at a Solution



I think that for the first question the answer is:

tanθ= µ_s → 11.31º

And for the rest I have no clue.
 
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Hi mmoadi! :smile:

Call the tension in the rope T, and apply good ol' Newtons's second law twice, once for each climber (and remember that you'll be using µs for one climber, and µk for the other). :wink:
 
First a question: For the first part "How much does the most inclination have to be for them not to slip?" when I got θ= 11.31º, is this correct?

So you mean that I have to put the two climbers into the equilibrium?

F_net(m1)= T+ m1gsinθ + µ_sm1gcosθ= ma
F_net(m2)= T- m2gsinθ - µ_km2gcosθ= ma

F_net(m1) = F_net(m2)
T+ m1gsinθ + µ_sm1gcosθ = T- m2gsinθ - µ_km2gcosθ

tanθ= (-µ_km2- µ_sm1) / (m1+m2)
θ= 9.46º
 
Last edited:
Hi mmoadi! :smile:

(try using the X2 tag just above the Reply box … it makes things a lot easier to read! :wink:)
mmoadi said:
First a question: For the first part "How much does the most inclination have to be for them not to slip?" when I got θ= 11.31º, is this correct?

Yes, that's fine. :smile:
T+ m1gsinθ + µ_sm1gcosθ = T- m2gsinθ - µ_km2gcosθ

Yes, that's fine too,

(I haven't checked the actual arithmetic)

except that you didn't need to split it into two first …

you could have used the net forces on the-two-climbers-and-the-rope as a single rigid body

… that gives you the same equation (but without the Ts) immediately, doesn't it? :wink:

(you need to use T when there are pulleys etc only because a pulley system as a whole isn't rigid)
 
That leaves us with the third part:
"-How much does the angle have to be if the roles were switched, the first climber slipping, and the second climber is trying to restrain him?"

Do I use the same approach or because the second climber is below the first climber the situation changes?
 
I have posted another problem. Would it be too much if I asked you to check it for me?
The problem is posted under:
A Banked Turn With Friction- min and max velocity
 
mmoadi said:
That leaves us with the third part:
"-How much does the angle have to be if the roles were switched, the first climber slipping, and the second climber is trying to restrain him?"

Do I use the same approach or because the second climber is below the first climber the situation changes?

yes , of course you use the same approach (just swap the masses) … why not? :confused:
 
Thank you very much! Have a nice evening!
 
Here I go again :)
So I swapped the masses and I got the final equation that looks like this:

tanθ= (-µ_km1- µ_sm2) / (m1+m2)
θ= 7.59º

Did I succeed?

Thank you for helping!
 
Last edited:
  • #10
I forgot to upload the picture.
Sorry, I posted picture on the wrong problem. Ooops!
It suppose to be attached to my other problem: How much ahead the front axle does the wheel jet the water that it pick
 

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  • #11
mmoadi said:
Here I go again :)
So I swapped the masses and I got the final equation that looks like this:

tanθ= (-µ_km1- µ_sm2) / (m1+m2)
θ= 7.59º

Did I succeed?

Again, I haven't checked the arithmetic, but yes the equation is fine. :smile:

(except, as in the previous equation, tanθ should obviously be positive … the error was in your original equations, where you had g and friction both going downhill! :rolleyes:)
 
  • #12
Thank you for all the help! Have a nice weekend!
 

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