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  • 2 years ago

A screw jack has 90 threads to the metre; the effort is applied at the end of an arm 30 cm long. What force must be applied to lift a load of 120 kg, if the efficiency at that load is 25% ?

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  1. whpalmer4
    • 2 years ago
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    F = (Q · p) / (2 · π · R) F is the force applied Q is the load p is thread pitch (distance for 1 turn) R is lever arm radius Here we have \(Q = 120 kg\), \(p = 1/90 m\), \(R = 0.3 m\) and our net force will be quadrupled because the efficiency is only 25%.

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