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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:
- Distance moved by the effort
- Work done by the effort
- M.A. of the pulley system
- Efficiency of the pulley system
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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 × dL
= 4 × 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) = `"Force"_"Effort" xx d_E`
WE = 50 × 60
= 3000 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 × 15
= 2625 kgf
:. η = `2625/3000`
= 0.875 × 100
= 87.5 %
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