Critical angle- two climbers in a roped party

AI Thread Summary
The discussion revolves around determining the critical angles for two climbers on a frosted slope to avoid slipping while connected by a rope. The first climber's angle of inclination to prevent slipping is calculated to be 11.31 degrees, while the second climber's angle, when roles are reversed, is found to be 7.59 degrees. Participants emphasize using Newton's laws and the coefficients of static and kinetic friction to derive these angles. There is a focus on ensuring the correct application of forces and the importance of treating the climbers as a single rigid body for calculations. Overall, the thread highlights the mathematical approach to solving the problem of climber safety on an inclined surface.
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
 
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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!
 
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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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