The handle of a water pump is $90 c m$ long from its piston rod. If the pivot handle is at a distance of $15 c m$ from the piston rod, calculate the least effort required at its other end to overcome a resistance of $60 k g f$ .

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Solution

Given:
Load arm = $15 c m$
length of water pump= $90 c m$
Effort arm = handle of water pump - load arm = $90 - 15 = 75 c m$

To calculate mechanical advantage:
Formula used :
$M A = \frac{E f f o r t a r m}{l o a d a r m} \Rightarrow M A = \frac{75}{15} = 5$

To calculate effort:
Thus, effort the required to overcome resistance or load of 60 kgf is,
Formula used: $M A = \frac{L o a d}{E f f o r t}$

Thus, $E f f o r t = \frac{l o a d}{M A} = \frac{60 k g f}{5} = 12 k g f$

Answer:
Hence the effort required is $12 k g f$ .

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

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