Answer to Question #186389 in Mechanics | Relativity for Damien

Question

A screw jack has a two-start thread of pitch 5mm. An effort of 40N is applied tangentially to the tommy bar, at a radius of 350 mm, to lift a load of 2200 N.

1. Calculate the velocity ratio and the mechanical advantage of the jack.
2. Calculate the work done in overcoming friction when the load is raised a distance of 75 mm.
3. Evaluate the efficiency of the screw jack and discuss what effect it has on the likelihood of overhauling and the law of the machine.

Answer 1

Pitch, $p=5\space mm=5\times10^{-3}\space m$

Radius, $r=350\space mm=350\times10^{-3}\space m$

Effort, $E=40\space N$

Load, $L=2200\space N$

(1) Velocity Ratio, $(V.R)=\dfrac{2\pi r}{p}=\dfrac{2\times3.14\times350\times10^{-3}}{5\times10^{-3}}=439.6$

Mechanical Advantage, $(MA)=\dfrac{Load}{Effort}=\dfrac{2200}{40}=55$

(2) Height raised, $h=75\space mm=75\times10^{-3}\space m$

Efficiency, $\eta=\dfrac{MA}{VR}=\dfrac{55}{439.6}\times100=12.511\%$

Work Done $=\dfrac{mgh}{\eta}=\dfrac{2200\times75\times10^{-3}}{0.125}=1320\space J$

(3) $Efficiency, \space \eta=12.511\%$