Question

A 15.0kg block is resting on a 2.5 m long board with a coefficient of static friction of 0.45. If one end of the board is slowly raised, at what angle will the block slip?

Answer 1

As the board is raised by an angle θ, the block will be held in place by 3 forces:

• its weight, pointing downward (magnitude w )

• the normal force of the board pushing up on the block, pointing in the direction perpendicular to the board (mag. n)

• static friction, pointing in the direction parallel to the board, opposite the direction the block would slip (mag. f )

Decompose the vectors into components that are parallel and perpendicular to the board (taking the direction of n to be the positive direction perpendicular to the board, and the direction of f to be the negative directoin parallel to it), so that by Newton's second law, we have

• net parallel force:

∑ F = w (//) + f

∑ F = m g sin(θ) - µ n = 0

where µ = 0.45 is the coefficient of static friction, g = 9.80 m/s² is the mag. of the acceleration due to gravity, and m = 15.0 kg is the mass of the block.

• net perpendicular force:

∑ F = n + w (⟂)

∑ F = n - m g cos(θ) = 0

Solve for n in the equation for net perpendicular force:

n = m g cos(θ)

Substitute this into the equation for net parallel force:

m g sin(θ) - 0.45 m g cos(θ) = 0

Solve for θ :

sin(θ) - 0.45 cos(θ) = 0

sin(θ) = 0.45 cos(θ)

tan(θ) = 0.45

θ = tan⁻¹(0.45)

θ ≈ 24.2°

So the maximum angle the board can be lifted before the block starts to slide is about 24.2°, since the coefficient of static friction µ is such that

f = µ n

where f is the maximum magnitude of the static friction force.