(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex]\Sigma[/tex]F_{y}=ma (sum of all forces equals to mass times acceleration, in case my symbols aren't same as the commonly used ones)

F_{fr}=[tex]\mu[/tex]F_{N}

F_{G}=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:

[tex]\Sigma[/tex]F_{y}=ma

F_{fr}- F_{G}= (63)(0)

[tex]\mu[/tex]F_{N}- mg = 0

[tex]\mu[/tex]F_{N}= mg

F_{N}= mg/[tex]\mu[/tex]

F_{N}= ((63)(9.8))/(0.83)

F_{N}= 744N

For 0.55

[tex]\Sigma[/tex]F_{y}=ma

F_{fr}- F_{G}= (63)(0)

[tex]\mu[/tex]F_{N}- mg = 0

[tex]\mu[/tex]F_{N}= mg

F_{N}= mg/[tex]\mu[/tex]

F_{N}= ((63)(9.8))/(0.55)

F_{N}= 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?

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# Homework Help: Normal Force Exerted

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