# Normal Force Exerted

1. Jan 29, 2009

### 312213

1. The problem statement, all variables and given/known data
The 63 kg climber in Fig. 4-52 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 0.83 and 0.55, 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 actual picture is of a man holding on to a rope with his back press against a right wall and his feet against a left wall.

2. Relevant equations
$$\Sigma$$Fy=ma (sum of all forces equals to mass times acceleration, in case my symbols aren't same as the commonly used ones)
Ffr=$$\mu$$FN
FG=mg (Gravity force equals mass times gravity)

3. The attempt at a solution
http://img301.imageshack.us/img301/6979/fbdyu9.jpg [Broken]

The diagram that I thought that could be it,though I doubt it.

For 0.83:
$$\Sigma$$Fy=ma
Ffr - FG = (63)(0)
$$\mu$$FN - mg = 0
$$\mu$$FN = mg
FN = mg/$$\mu$$
FN = ((63)(9.8))/(0.83)
FN = 744N

For 0.55
$$\Sigma$$Fy=ma
Ffr - FG = (63)(0)
$$\mu$$FN - mg = 0
$$\mu$$FN = mg
FN = mg/$$\mu$$
FN = ((63)(9.8))/(0.55)
FN = 1123N

With these two numbers, I added for 1866.4N (using the two numbers but with more decimal places). I also subtracted 1123 by 744 for 378.7N.

Both ways were wrong.

I lack any confidence that any of this was done correctly. First off, how is the diagram supposed to be? The way my diagram is seems to lack reasoning, unless the two diagrams are combined. Since I am not sure if my diagram is right, I doubt my net equation is right either. Could some one help my understand how the correct diagram is?

Last edited by a moderator: May 3, 2017
2. Jan 29, 2009

### Delphi51

Your diagram is pretty good - the same normal force on both sides, the two friction forces upward. But you should only have ONE force of gravity downward. Combine the diagram or at least the equations so you have only one sum of forces equaling zero. You will be able to factor out the FN and solve for it.

3. Jan 29, 2009

### 312213

I don't really understand what one force of gravity would mean. The way I understand that to be is that one of my side is left alone and the other one has 0 for Fg but that would result that whole side to equal zero (FN = ((63)(0))/(0.55) = 0) so I must be misunderstanding this.

Another thing is that is am I supposed to do a $$\Sigma$$Fx=ma?

4. Jan 29, 2009

### Delphi51

The thing is the guy has only one force of gravity acting on him. One weight. If you want to get really sophisticated, that force would be at his center of mass and there might be a problem of one end of him falling while the other is secure. But don't worry about that!

Yes, use sum of forces equals ma which equals zero. You'll have mg down and the two friction forces up. Solve for Fn.