Static friction on inclined surface

AI Thread Summary
The discussion revolves around calculating the coefficient of static friction for a boy on a 56-degree slope, given his mass of 52 kg and a cord with a breaking strength of 152 Newtons. The forces acting on the boy include the gravitational force down the slope and the normal force, which are expressed as Fd = mg sin(56°) and N = mg cos(56°), respectively. The frictional force is represented as Fr = μN, and the maximum force from the string is Fs = 152 N. The key equation derived is Fd ≤ μN + Fs, indicating that the downward force must not exceed the combined forces of friction and the string's tension. The discussion concludes that as long as the coefficient of friction (μ) is sufficiently high, the boy will remain secure on the slope.
drkidd22
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Hello all.

I'm having issues trying to figure out the following problem.

A boy with a mass of 52 kg and on a 56-degree slope is hanging on to a cord with a breaking strength of 152 Newtons. What must be the coefficient of static friction between the boy and the surface?
 
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force down the slope:
Fd = mg sin 560

normal force:
N = mg cos 560

Friction holding the boy
Fr = \muN

maximum force provided by string
Fs = 152

Equating forces:
the string must not break so the boy doesn't fall, so
Fd \leq \muN + Fs

you can the solve for \mu, the coefficient of friction

Hope this is correct and helps
 
This works well. I had gotten almost everything together. The only thing I couldn't figure out was the equation of the forces.

Thanks for your help.
 
well, i was just being a little bit fancy

if you want, i think you can use

Fd = \muN + Fs

the previous equation:
Fd \leq \muN + Fs
just means that the force pulling the boy down must be less than or equal to the force holding the boy up (Fr and Fs)
The only unknown variable is \mu, so as long as \mu is big enough so that friction force is big, the boy would not fall.
 
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