# Static Friction of a climber

1. Aug 26, 2009

### keropi080

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

The 70-kg climber is supported in the "chimney" by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that friction forces are both at a maximum.

The diagram shows the climber with his back against one wall, his shoes on another wall, and clinging on a rope between those two walls.

g = 9.8 m/s^2
m = 70kg
Fs (coefficients) = 0.80; 0.60

2. Relevant equations

Fnet = ma
Fgrav = mg
Fs = (myu)(Fn)

3. The attempt at a solution

I drew the FBD with the friction forces going down, making sure that the two friction forces + Fgrav is equal to the tension force + the normal force he exerts, which means that it's not moving.

So...
Ff(shoes) + Ff(back) + mg = Ft + Fn
Fn = myu*Fn(shoes) + myu*Fn(back) + (70kg)(9.8m/s^2) - Ft
Fn = 0.80*Fn(shoes) + 0.60*Fn(back) + 686N - Ft

Now I'm stuck. :(

Last edited: Aug 26, 2009
2. Aug 26, 2009

### Fightfish

Firstly, the normal force exerted on him by the wall is not to be considered in the vertical equilibrium - it acts horizontally in this case!
Next, in what direction do the frictional forces on him act?

3. Aug 26, 2009

### keropi080

On the opposite walls? But then, the normal force on opposite sides would have to be equal, since he's not moving in the x-direction?

4. Aug 26, 2009

### Fightfish

Kinda edited my earlier reply a bit upon realising something else.
Yup, that observation on the equality of the normal forces is correct. Regarding tension, what do you think the value is? Is he supported by the rope?

5. Aug 26, 2009

### kuruman

One error I see is adding the tension and the normal force. If I understand the problem correctly, they are in perpendicular directions. Look at your FBD and add the x and y components of the forces separately.

6. Aug 26, 2009

### keropi080

Let's see. Since both normal forces are equal...

Fn = Fnx(0.80+0.60) + 686N - Ft
Fn = Fnx(1.40) + 686N - Ft

I can't cancel out Fn because, as you said, they're not the same. How do I get the values of the remaining variables?

7. Aug 26, 2009

### keropi080

Maybe I misunderstood the question...
He doesn't necessarily have to move up? So that means, the normal force being asked on the question are the forces he exerts on his feet and back? Is that right?

8. Aug 26, 2009

### Fightfish

I really need to stop editing my posts and write everything coherently at one go lol :p
Start over again in forming your force equilibrium equation.
Problems in Ff(shoes) + Ff(back) + mg = Ft + Fn:
1. In what direction do the frictional forces on him act?
2. Read the question; does the rope support him?
3. The normal force exerted on him is in the horizontal direction and should not appear in the static equilibrium consideration in the vertical direction. Consider separately.

9. Aug 26, 2009

### keropi080

1.) Along the x-axis, on the walls.
2.) The friction forces are at their maximum, which means that whichever direction he's going, up or down, he's being supported by his feet and back! Even if he lets go of the rope, he's still in equilibrium! Wow, big realization. :O

10. Aug 26, 2009

### Fightfish

1. Along the x-axis is not clear enough; the frictional forces must act upwards to support his weight - but in your equation you seemed to indicate that it is acting downwards.

2. The question states "The 70-kg climber is supported in the "chimney" by the friction forces exerted on his shoes and back." The rope can only support the climber if it is taut. When the rope of a climber is taut, that can only mean one thing - he must have fallen already.

11. Aug 26, 2009

### keropi080

I get it now! Thank you so much!
I didn't what that first sentence meant. I overlooked it. Thanks again!