2. A load of 1.5 kN, resting on an inclined rough plane, can be moved up the plane by a force of
2 kN applied horizontally or by a force 1.25 kN applied parallel to the plane. Find the
inclination of the plane and the coefficient of friction.
Answers
Answer:
A load of 150 kn resting on an inclined plane can be moved up the plane by a force of 2kn, applied horizontally or by a force of 125 kn applied parallel to the plane. Can you find the inclination of the plane and the coefficient of friction?
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A load of 150 kn resting on an inclined plane can be moved up the plane by a force of 2kn, applied horizontally or by a force of 125 kn applied parallel to the plane. Can you find the inclination of the plane and the coefficient of friction?
The force parallel to the plane is Fp = 150Sinθ, and the force normal to the plane Fn = 150Cosθ.
The given information states that Fp = 125kN, so 150Sinθ = 125, or
Sinθ = 125/150 = 0.8333, in which case, θ = 56.44°
BUT the coefficient of friction μ = Fp/Fn = 150Sinθ/150Cosθ = Tanθ
So μ = Tan56.44° = 1.507, which means that Fp > Fn
This whole question is nonsense – especially the part about the object being moved by a force of 2kN applied horizontally.
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