A pulley system with V.R. = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction. (g = 10 N kg-1)
Calculate:
(i) Distance moved by the effort.
(ii) Work done by the effort.
(iii) M.A. of the pulley system.
(iv) Efficiency of the pulley system.
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Given,
V.R. = 4
L = 175 kgf
dL = 15 m
E = 50 kgf
(i) V.R. = $\frac{d_E}{d_L}$
Substituting the values we get,
$$\begin{aligned}4 = \frac{d_E}{15} \\[0.5em] \Rightarrow d_E = 15 \times 4 \\[0.5em] \Rightarrow d_E = 60 \text{ m}\end{aligned}$$
(ii) Work done by the effort = E x dE
= 50 x 10 x 60
= 3 x 104
(iii) M.A. = $\frac{L}{E}$
Substituting the values we get,
$\text {M.A.} = \frac{175}{50} = 3.5$
(iv) Efficiency η = $\frac{M.A.}{V.R.}$
Substituting the values we get,
$$\begin{aligned}η = \frac{3.5}{4} = 0.875 \\[0.5em] \therefore η = 87.5 %\end{aligned}$$