weight component down = m g cos 34
friction force up = m g cos 34 * mu
so
mu m g cos 34 + 240 = m g sin 34
HELP!!!
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.
4 answers
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.
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.