posted by Randy
A box rest on an incline making a 34 angle with the horizontal. It is found that a parallel force to the incline of at least 240 N can prevent the box from sliding down the incline. If the weight of the box is 800 N, find the coefficient of static friction between the box and the incline.
weight component down = m g cos 34
friction force up = m g cos 34 * mu
mu m g cos 34 + 240 = m g sin 34
I get 8.01. That doesn't seem correct. I plugged in all my numbers. Anything you think I could have done incorrectly?
I converted newtons to kilograms and my answer makes more sense to me. I got 0.312 for mu this time. Thank you for the guidance.
M*g = 800 N.
Fp = 800*sin34 = 447.4 N.
Fn = 800*Cos34 = 663.2 N.
Fap-Fp-Fs = M*a.
240-447.4 + Fs = M*0 = 0
Fs = 207.4 N. = Force of static friction.
u = Fs/Fn = 207.4/663.2 = 0.313.