A 50-kg block is being pulled up a 30.0^ slope by force of 250N which is parallel to the slope, but the block does not slide up the slope. What is the minimum value of the coefficient of static friction required for this to happen?

Answer 1

M*g = 50 * 9.8 = 490 N. = Wt. of block. = Normal force(Fn).

Fp = 490*sin30 = 245 N. = Force parallel with the incline.

Fs = u*Fn = u * 490 = 490u.

Fap-Fp-uFn = M*a.
250 - 245 - 490u = M*0 = 0,
u = 0.010.