the handle of the screw jack is 35cm long and the pitch of the screw is 0.5cm.what force must be applied at the end of the handle when lifting a load of 2200N if the efficiency of the jack is 40%
goes up 0.5cm
in distance of 2 pi r = 2 pi * 35 = 220
so mechanical advantage = 220/ 0.5 = 440
ideally 2200 N / 440 = 0.5N
but 0.40 efficient so
0.5 / 0.40 = 1.25 N