Critical angle- two climbers in a roped party

In summary, two climbers are climbing up a frosted hillside in a roped party. The most inclination must be 11.31º for them not to slip, given a coefficient of kinetic friction of 0.1 and a coefficient of static friction of 0.2. If the first climber starts to slip, the most inclination for the second climber to restrain the fall is 9.46º. And if the roles were switched, the first climber slipping and the second climber trying to restrain, the most inclination would be 7.59º.
  • #1
mmoadi
157
0

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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  • #2
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:
 
  • #3
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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  • #4
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)
 
  • #5
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
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
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:
 
  • #8
Thank you very much! Have a nice evening!
 
  • #9
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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  • #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!
 

Related to Critical angle- two climbers in a roped party

1. What is the critical angle in rock climbing?

The critical angle in rock climbing refers to the maximum angle of the rock face where a climber can maintain their grip and prevent a fall without the use of additional support, such as ropes or gear.

2. How is the critical angle determined?

The critical angle is determined by factors such as the type and texture of the rock, the angle of the rock face, and the skill and strength of the climber. It can also vary depending on weather conditions and the route being climbed.

3. Why is the critical angle important for climbers?

The critical angle is important for climbers because it determines the level of difficulty and risk involved in a climb. Climbers must be able to assess and navigate around critical angles in order to reach their desired destination and avoid potential falls.

4. What is a roped party in rock climbing?

A roped party in rock climbing refers to a group of climbers who are connected by a rope and are climbing together as a team. This is a common practice for safety purposes, as it allows for protection and assistance in the event of a fall.

5. How does the critical angle affect a roped party?

The critical angle can greatly affect a roped party in rock climbing. If the critical angle is too steep, it may be difficult for the climbers to maintain their grip and progress upwards. Additionally, if one member of the roped party falls, the critical angle can determine the likelihood of the others being pulled down with them.

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