The height of the inclined plane is 1.2 m, and the length is 10.8 m. To lift a load weighing 180 kg on this inclined

| June 20, 2021 | education

The height of the inclined plane is 1.2 m, and the length is 10.8 m. To lift a load weighing 180 kg on this inclined plane, a force of 250 N was required. Determine the efficiency of the inclined plane and the friction force.

h = 1.2 m.
S = 10.8 m.
m = 180 kg.
g = 9.8 m / s ^ 2.
F = 250 N.
Ftr -?
Efficiency -?
The efficiency is determined by the formula: efficiency = Apol * 100% / Azatr, where Apol is useful work, Azat is work expended.
Apol = m * g * h.
Azatr = F * S.
Efficiency = m * g * h * 100% / F * S.
Efficiency = 180 kg * 9.8 m / s ^ 2 * 1.2 m * 100% / 250 N * 10.8 m = 78.4%.
Let us find the work of the friction force with uniform lifting along the inclined plane: F = Ffr + m * g * sinα, ∠α is the angle of the inclined plane.
sinα = h / S.
F = Ftr + m * g * h / S.
Ftr = F – m * g * h / S.
Ftr = 250 N – 180 kg * 9.8 m / s ^ 2 * 1.2 m / 10.8 m = 54 N.
Answer: efficiency = 78.4%, while lifting, the friction force Ffr = 54 N. acts.



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