A box of mass 50kg is placed on an inclined plane. When the angle of the plane is increased to 30°, the box begins to slide downward. Calculate the coefficient of statics friction b/w the plane and box . Draw the free body diagram .

Answer 1

M*g = 50 * 9.8 = 490 N. = Wt. of box.

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

Fn = 490*Cos30 = 424 N. = Normal force.

Fs = u*Fn = = 424u. = Force of static friction.

Fp-424u = M*a.
245-424u = M*0 = 0, U = ?.

Answer 2

20.6