Question

a concrete brick is placed on a polished wood board which is initially laying horizontally on the floor. the static coefficient of friction is 0.63 between the board and the brick. the kinetic coefficient of friction is 0.55 between the two surfaces. we gradually raise one end of the wood board so that the board is tilted to increasingly larger angles, θ, relative to the horizontal. calculate the critical angle, θcrit at which the brick will start to slide along the wood board.

Answer 1

Static friction holds the brick in place up until the board is raised to the critical angle, at which point static friction is maximum. The net forces on the brick parallel and perpendicular to the board are

∑ F[para] = mg sin(θ) - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(θ) = 0

where mg is the weight of the brick.

It follows that

F[normal] = mg cos(θ)

F[friction] = 0.63 F[normal] = 0.63 mg cos(θ)

mg sin(θ) - 0.63 mg cos(θ) = 0

Solve for θ :

sin(θ) - 0.63 cos(θ) = 0

sin(θ) = 0.63 cos(θ)

sin(θ)/cos(θ) = 0.63

tan(θ) = 0.63

θ = arctan(0.63) ≈ 32°