Question

A basket of flowers weighing 2kg is placed on a flat grassy slope. The coefficient of static friction between the basket and the slope is 0.4, and the basket is on the point of slipping down the slope.

Draw a force diagram to indicate forces (including friction) that act on the basket.

Write down vector expressions for the forces and the equilibrium condition. Hence determine the angle that the slope makes with the horizontal.

Really not sure where to even start

Answer 1

M*g = 2 * 9.8 = 19.6 N. = Wt. of flowers.

Fp = 19.6*sin A. = Force parallel with incline.

Fn = 19.6*Cos A. = Normal force.

Fs = u*Fn = 0.4*19.6*Cos A = 7.84*Cos A. = Force of static friction.

Fp-Fs = M*a.
19.6*sin A-7.84*Cos A = M*0 = 0,
19.6*sin A = 7.84Cos A,
19.6*sin A/Cos A = 7.84,
sin A/Cos A/Cos A = 7.84/19.6 = 0.4,
sin A/Cos A = Tan A,
Tan A = 0.4, A = 21.8o.