# Critical angle- two climbers in a roped party

1. Oct 21, 2009

1. The problem statement, all variables and given/known data

-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.

2. Relevant equations

tanθ= µ_s

3. 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.

2. Oct 21, 2009

### tiny-tim

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).

3. Oct 22, 2009

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: Oct 23, 2009
4. Oct 22, 2009

### tiny-tim

(try using the X2 tag just above the Reply box … it makes things a lot easier to read! )
Yes, that's fine.
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?

(you need to use T when there are pulleys etc only because a pulley system as a whole isn't rigid)

5. Oct 22, 2009

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?

6. Oct 22, 2009

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

7. Oct 22, 2009

### tiny-tim

yes , of course you use the same approach (just swap the masses) … why not?

8. Oct 22, 2009

Thank you very much! Have a nice evening!

9. Oct 23, 2009

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: Oct 23, 2009
10. Oct 23, 2009

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. Oct 23, 2009

### tiny-tim

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

(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! )

12. Oct 23, 2009