The force F 1 parallel to inclined plane that is necessary to move a body up an inclined plane is double the force F 2 that is necessary to just prevent it from sliding down, then:
Where φ = Limiting angle of repose, θ = angle of inclined plane, w = weight of the body
Text Solution
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(a, d)
F 1 = mg sin θ + μ mg cos θ .
F 2 = mg sin θ – μ mg cos θ .
But mg = w
μ = tan φ
∴ F 1 = w (sin θ + 
cos θ ) ⇒ F 1 = w sin ( θ + φ ) sec φ
∴ Now F 1 = 2 F 2
mg sin θ + μ mg cos θ = 2 (mg sin θ – μ mg cos θ )
sin θ + μ cos θ = 2 sin θ – 2 μ cos θ ⇒ 3 μ cos θ = sin θ ⇒ tan θ = 3 μ
tan θ = 3tan φ .
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