Question

The pitch of a jackscrew shown in figure 2 is 0.3 cm and has a lever arm of 60 cm length.

What force at the end of the lever arm will raise a load of 190 kg if the efficiency of the machine

is 12 %?

Answer 1

The ratio of output force to input force of a jackscrew is:

$\frac{F_2}{F_1}=\eta \frac{l_1}{l_2}$ ,

where $\eta$ - the efficiency of the machine,

$l_1$ - a lever arm,

$l_2$ - the pitch of a jackscrew.

So,

$F_1=\frac{F_2l_2}{\eta l_1}=\frac{mgl_2}{\eta l_1}=\frac{190\cdot 9.8\cdot 0.3}{0.12\cdot 60}\approx77.6\space N$ .