Question

a screw jack with pitch 0.5cm and the handle of the screw jack is 2cm long.Determine the force(effort)that must be applied at the end when lifting a load of 200N.Consider the efficiency of the screw to be 35%.

Answer 1

One complete rotation of the handle of the screw jack leads to 0.5 cm vertical displacement of the load. So, the velocity ratio is

$$VR=\frac{2\pi r}{p}.$$

The mechanical advantage is

$$MA=\eta VR=\frac{2\pi\eta r}{p}.$$

So, the applied force is

$$F_a=\frac{F}{MA}=\frac{Fp}{2\pi\eta r},\\\space\\ F_a=\frac{200·0.005}{2\pi·0.35·0.02}=22.7\text{ N}.$$