Question

with a machine of 65% efficiency if one raises a load of 600n through a distance of 15m applying a force of effort which is equal to 600n calculate the ima(vr) of machine and the distance that an effort moved

Answer 1

Explanations & Calculations

• The efficiency of a machine is given by

$\qquad\qquad \begin{aligned} \small Eff &=\small \frac{Input \, work}{Output \, work}\times 100\\ \end{aligned}$

• Work output is

$\qquad\qquad \begin{aligned} \small &=\small Fs=600\,N\times 15m=9000J \end{aligned}$

• Work input if the distance effort travels is some $x \,(m)$,

$\qquad\qquad \begin{aligned} \small &=\small 600N\times x=600x \end{aligned}$

• Then by the definition

$\qquad\qquad \begin{aligned} \small 65&=\small \frac{9000}{600x}\times 100\\ \small x &=\small \bold{23.08\,m}\\ \end{aligned}$

• The ideal mechanical advantage can be calculated to be

$\qquad\qquad \begin{aligned} \small IMA&=\small \frac{distance_{effort}}{distance_{load}}=\frac{23.08m}{15m}\\ &=\small \bold{1.539} \end{aligned}$