Applying a force of 2.5 kN to the long arm of the lever, they lifted a load weighing 1 ton, suspended

| August 13, 2021 | education

Applying a force of 2.5 kN to the long arm of the lever, they lifted a load weighing 1 ton, suspended on the short arm of the lever. The load was raised to a height of 0.8 m, while the point of application of the driving force dropped to a height of 4 m. Determine Lever Efficiency

Data: F (force on the long arm of the lever) = 2.5 kN (2500 N); m (mass of the lifted load) = 1 t (in SI m = 1000 kg); h1 (lifting height) = 0.8 m; h2 (height of lowering of the point of application of force) = 4 m.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

To determine the efficiency of the used lever, we apply the formula: η = Ap / Az = m * g * h1 / (F * h2).

Let’s calculate: η = 1000 * 10 * 0.8 / (2500 * 4) = 0.8 (80%).

Answer: The efficiency of the lever is 80%.



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