Question

A block is at rest on an inclined plane.
63 kg µs = 0.47
èc
Find the critical angle, èc, at which the block just begins to slide. Answer in units of ◦

Answer 1

M*g = 63 * 9.8 = 617.4 N. = Wt. of block

Fp = Fs @ critical angle.
Mg*sinA = us*Mg*CosA
Divide both sides by Mg:
sin A = us*Cos A
Divide both sides by Cos A:
sin A/Cos A = us = 0.47 = Tan A
A = 25.2o = The critical angle.

Fp = 262.6

Answer 2

θ = tan⁻¹μ