Ablock of mass m = 3 kg is held against a wall with a force f applied at an angle φ, as shown. there is a coefficient of static friction μs = 0.25 between the block and the wall. if φ = 60°, then the minimum f required to hold the block up is

29.7 N

Explanation:

At the minimum force F, the block will just begin to slide down, so friction force will be pointing up.

Sum of the forces in the x direction:

∑F = ma

N − F cos φ = 0

N = F cos φ

Sum of the forces in the y direction:

∑F = ma

Nμ + F sin φ − mg = 0

Substitute:

(F cos φ)μ + F sin φ − mg = 0

F (μ cos φ + sin φ) = mg

F = mg / (μ cos φ + sin φ)

Given m = 3 kg, μ = 0.25, and φ = 60°:

F = (3)(9.81) / (0.25 cos 60° + sin 60°)

F ≈ 29.7 N

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