Question

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

Answer 1

Answer:

(b) mechanical advantage of handle

Explanation:

The handle of a water pump is 90 cm long from its piston rod if the pivot of the handle is at a...

Answer 2

Answer:

Let's first calculate mechanical advantage

load arm = 15 cm

Effort arm = handle of water pump - load arm

= 90-15=75cm

$ma = \frac{effort \: arm}{load \: arm} = \frac{75}{15} = 5$

Thus effort required to overcome resistance or load of 60 kgf is ,

$ma = \frac{load}{effort}$

Thus ,

$effort = \frac{load}{ma} = \frac{60kgf}{5} = 5$

Thus effort= 12 kgf

Explanation:

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