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Question
A boy lifts a load of 40 kgf through a vertical height of 2m in 5s by using a single fixed pulley when he applies an effort of 48 kgf. Calculate:
(i) the mechanical advantage, and
(ii) the efficiency of the pulley. Why is the efficiency of the pulley is not 100%?
(iii) the energy gained by the load in 5s, and
(iv) the power developed by the boy in raising the load.
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Solution
Given:
L = 40 kgf,
dL = 2m,
t = 5 s,
E = 48 Kgf.
(i) Mechanical advantage (M.A.) =`"L"/"E"=40/48=5/6=0.833`
(ii) If the effort moves a distance d downwards, the load also moves a distance d upwards. So velocity ratio (V.R.) = d/d = 1,
Efficiency = `"M.A."/"V.R."=0.833/1`
= 0.833 (or 83.3%).
The efficiency of the pulley is not 100% because some energy is wasted in overcoming the friction in the pulley bearings.
(iii) The energy gained by the load in 5s = Load × Displacement of load in 5s
= 40 kgf × 2m
= 80 kgf × m.
(iv) Power developed by the boy =`"Effort×Displacement of effort"/"Time"=(48"kgf"×2"m")/(5"s")`
= 19.2 kgf × ms−1.
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