A handle of a water pump is $90 c m$ long from its piston rod. If the pivot of the 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 $60 k g f$ ? and what is the mechanical advantage?

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Solution

Step 1: Given data
Load arm $= 15 c m .$
Effort arm $= T o t a l l e n g t h o f p i s t o n - L o a d a r m = 90 - 15 = 75 c m .$
Load $= 60 k g f .$
Step 2: Formula
Mechanical advantage(MA)
$M A = \frac{L o a d}{E f f o r t} = \frac{E f f o r t a r m}{L o a d a r m} M A = \frac{E f f o r t a r m}{L o a d a r m} = \frac{75}{15} = 5$
Step 3: Calculate mechanical advantage
We know that,
Mechanical advantage = $\frac{L o a d}{E f f o r t}$
Effort = $\frac{L o a d}{M A}$
Effort = $\frac{60}{5} = 12 k g f$
Hence, the effort required is $12 k g f$ and the mechanical advantage is $5 .$

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A handle of a water pump is 90cm long from its piston rod. If the pivot of the handle is at a distance of 15cm from the piston rod, calculate the least effort required at its other end to overcome a resistance of60kgf? and what is the mechanical advantage?

A handle of a water pump is $90 c m$ long from its piston rod. If the pivot of the 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 $60 k g f$ ? and what is the mechanical advantage?