Question

A fixed pulley is driven by a $100kgf$ mass falling at a rate of $8.0m$ in $4.0s$. It lifts a load of $75.0kgf$. Calculate the efficiency of the pulley and the height to which the load is raised in $4.0s$.

• A. $0.65,4.0m$
• B. $0.85,8.0m$
• C. $0.75,4.0m$
• D. $0.75,8.0m$

Answer 1

Answer: (D)

$0.75,8.0m$

As given above effort $\left(E\right)=100kgf=10kg$
Load$\left(L\right)=75kgf=7.5kg$
$M.A=$$\frac{L}{E}$$=\frac{7.5}{10}$$=0.75$
From the definition of velocity ratio, there is only single pulley. Hence, it will be $\frac{d_e}{d_l/n}$
where, $n$=number of pulley
$d_e$=distance moved by effort
$d_l$ = distance moved by load
So velocity of effort =$\frac{8}{4}$$=2$
Velocity of load=$\frac{x}{4}$
Hence equating it with velocity ratio we find,
1 = $\frac{2}{x/4}$
$x=8m$ is the height at which load is raised
So, $V.R=1$
Efficiency, $\eta$ = $\frac{M.A}{V.R}\times 100=$$\frac{0.75}{1}\times 100=75$%