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?

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.

answer; equivalent to a one(1) horsepower motor;

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