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The handle of a water pump is 90 cm long from its piston rod. If the pivot handle is at a distance of 15 cm from the piston rod, calculate (a) least effort required at its other end to overcome a resistance of 60 kgf and (b) mechanical advantage of handle.
Asked by ploganathan9 | 18 Dec, 2019, 07:45: PM
Effort arm = handle of water pump - load arm = 90 - 15 = 75 cm
MA = Effort arm/load arm = 75/15 = 5

Thus, effort required to overcome resistance or load of 60 kgf is,
Thus,
Effort = load/ MA = 60 kgf/5 = 12 kgf

Thus, effort = 12 kgf
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