Question

A pulley system as shown in figure 4 has an ideal mechanical advantage of 5 and is used to

lift a load of 1000 lb. If the effort moves 10 ft, how far does the load move? If this work was

done in 15 s, what horsepower was developed? If the machine was 80 % efficient, how much

effort would be needed to lift the load?

Answer 1

Mechanical Effort $=\dfrac{Load}{Effort}$

$\Rightarrow Effort=\dfrac{1000}{5}=200\space lb$

Height moved by effort = $\dfrac{10}{5}=2ft$

Work done = $mgh=1000\times32.17\times2=64340\space ft.lb$

Power, $P=\dfrac{Work\space done}{time}=\dfrac{64340}{15}=4289.33\space ft.lb/s=7.79hp\space\space\space\space\space\space\space\space\space\space(\because\space1hp=550ft.lb/s)$

Effort needed $= \dfrac{m}{\eta M.A}=250\space lb$