A woman draws water from a well using a fixed pulley. The mass of bucket and water together is 6.0 kg the force applied by the woman is 70 N. Calculate the mechanical advantage. (g = 10 ms $^{- 2}$ )

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Solution

Given load L $= m g = 6 \times 10 = 60 N$ and effort $E = 70 N$
Now mechanical advantage
$= \frac{l o a d}{e f f o r t} = \frac{60 N}{70 N} = 0.86$

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A woman draws water from a well using a fixed pulley. The mass of bucket and water together is 6.0 kg the force applied by the woman is 70 N. Calculate the mechanical advantage. (g = 10 ms−2)

Given load L =mg=6×10=60N and effort E=70NNow mechanical advantage=loadeffort=60N70N=0.86

A woman draws water from a well using a fixed pulley. The mass of bucket and water together is 6.0 kg the force applied by the woman is 70 N. Calculate the mechanical advantage. (g = 10 ms $^{- 2}$ )

Given load L $= m g = 6 \times 10 = 60 N$ and effort $E = 70 N$
Now mechanical advantage
$= \frac{l o a d}{e f f o r t} = \frac{60 N}{70 N} = 0.86$