Asked by Jay

You have a mass of 69 kg and are on a 47 degree slope hanging on to a cord with a breaking strength of 160 newtons. What must be the coefficient of static friction to 2 decimal places between you and the surface for you to be saved from the fire?
Answered by Henry
Wt. = m*g = 69kg * 9.8N/kg = 676.2 N.

F1 = 676.2*sin47 = 494.54 N. = Force parallel to the slope.
F2 = 676.2*cos47 = 461.2 N. = Force perpendicular to the slope.

160-Fs = = m*a
160-u*F2 = m*0 = 0
u*461.2 = 160
u = 0.35 = Coefficient of static friction.
Answered by Tiff
Actually, Henry is incorrect. The acceleration is NOT equal to 0. The acceleration is F1/m thus, the equation he wrote (160 - Fs = m • a) should have m • a = F1.

Final equation:

Ft - uF2 = F1
160 - u(461.2) = 494.54
u = .725
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