Question

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% ?

Answer 1

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%.