Example involving Friction (continued)

Next assume the block tends to slide down the plane and construct a

FBD. Note the directions taken for + x and + y. Then write the equations of

equilibrium.

ΣF_y = 0 N ˗ P sinθ ˗ Q cosθ = 0

ΣF_x = 0 F + P cosθ ˗ Q sinθ = 0

So from ΣF_y = 0 N = P sinθ + Q cosθ and F_max = μ_s N

And putting this expression for F_max into ΣF_x = 0 and solving for P gives

P = Q ( sinθ - μ_s cosθ ) / (cosθ + μ_s sinθ )

And putting in the values for Q, θ, and μ_s gives P = 143 N.

So equilibrium of the block is maintained for

143 ≤ P ≤ 483 Newtons

i.e. Motion of the block impends downward if P = 143 N

and Motion of the block impends upward if P = 483 N

Click here for another example.

Return to Notes on Statics

Copyright © 2019 Richard C. Coddington
All rights reserved.