021 (part 1 of 3) 10.0 points A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.62 and µk = 0.53,

Question

respectively. The acceleration of gravity is 9.8 m/s2 .
36 kg µ
29◦
What is the frictional force acting on the 36 kg mass? Answer in units of N.
022 (part 2 of 3) 10.0 points What is the largest angle which the incline can have so that the mass does not slide down the incline? Answer in units of ◦.
023 (part 3 of 3) 10.0 points What is the acceleration of the block down the incline if the angle of the incline is 36◦ ? Answer in units of m/s2.

Answer 1

M*g = 36 * 9.8 = 352.8 N. = Wt. of block.

Fp = 352.8*sin29o = 171 N. = Force parallel to the incline.

Fn = 352.8*Cos29 = 308.6 N. = Normal force.

1. Fs = u*Fn = 0.62 * 308.6 = 191.3 N. = Force of static friction.

2. Fp = Fs.
352.8*sin A = 191.3
A = ?.

3. Fp = 352.8*sin36 = 207.4 N.

Fn = 352.8*Cos36 = 285.4 N.

Fk = u*Fn = 0.53 * 285.4 N. = 151.3 N. = Force of kinetic friction.

Fp-Fk = M*a.
207.4-151.3 = 36*a
a = ?.