Question

In a test on machine an effort of 50N was required to raise a load pf 250N. If the effort moves through 150 mm to raise the load by 25 mm, find the following:

i. The Mechanical Advantage

ii. The Velocity Ratio

iii. The efficiency

iv. The work done by the effort.

v. The work done in raising the load 25 mm.

vi. The power developed by the effort over a period of 5 seconds

Answer 1

Effort, $F_e=50\space N$

Load, $F_l=250\space N$

Distance moved by effort, $d_e=150\space mm=150\times10^{-3}\space m$

Distance moved by load, $d_l=25\space mm=25\times10^{-3} \space m$

(i) Mechanical Advantage, $(MA)=\dfrac{F_l}{F_e}=5$

(ii) Velocity Ratio, $(VR)=\dfrac{d_e}{d_l}=6$

(iii) Efficiency, $\eta=\dfrac{F_ld_l}{F_e d_e}=0.83$

(iv) Work done by effort, $W_e=F_ed_e=7.5\space J$

(v) Work done in raising the load, $W_l=F_ld_l=6.25\space J$

(vi) Power developed by the effort over a period of 5 seconds = $\dfrac{W_e}{t}$

$=\dfrac{7.5}{5}=1.5\space W$