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Question
A pulley system with VR = 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
1) Distance moved by the effort
2) Work done by the effort
3) M.A. of the pulley system
4) Efficiency of the pulley system
Solution
1) Displacement due to the effort is VR = 4
Load (L)= 175 kgf
Effort (E)= 50 kgf
Load displacement (dL)=15 m
`VR = d_E/d_L`
Effort displacement (dE) = 4 x dL = 4 x 15 = 60 m
Distance moved by the effort is 60 m
2) Work done by the effort is calculated using the following formula.
Work done by effort (WE)= ForceEffort x dE
WE = 50 x 60 = 30000 J
3) Mechanical advantage of the pulley system is calculated as
MA of the pulley system = `"Load (L)"/"Effort (E)"` = `175/50 = 3.5`
4) Efficiency of the pulley is the ratio of work output to the work input required by the system
Efficiency of the pulley system (η) = `"Work output (Work Done by Load)"/"Work input (Work Done by Effort)"`
Work done by load = ForceLoad x dL = 175 x 15 = 2625 kgf
:. η = `2625/3000` = 0.875 x 100 = 87.5 %
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