Question

The handle of a water pump is 90 cm long from its piston rod . If the pivot of handle is at a distance of 15 cm from piston rod , calculate a) least effort required at its other end to overcome a resistance of 60 kgf and b ) mechanical advantage of handle .

Sixth Standard . Numericals ​

Answer 1

Answer:

Let's first calculate mechanical advantage.

Load arm =15 cm  

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,  

MA = Load/Effort

Thus,  

Effort = load/ MA = 60 kgf/5 = 12 kgf  

 

Thus, effort = 12 kgf

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