A woman from a water from a well using a fixed pulley. The mass of the bucket and water together is 6 kg. The force applied by the woman is 70 N. Calculate the Mechanical advantage.(given g=9.8m/s2)

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Solution

Dear Student,
Mass of the bucket + water = M = 6 kg.
Weight of the bucket + water $= M g = 6 k g \times 9.8 m s^{- 2} = 58.8 N$
$M e c h a n i c a l A d v a n t a g e (M A) = \frac{F o r c e o u t p u t}{F o r c e i n p u t} = \frac{L o a d}{E f f o r t} = \frac{58.8}{70} = 0.84$
Regards

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A woman from a water from a well using a fixed pulley. The mass of the bucket and water together is 6 kg. The force applied by the woman is 70 N. Calculate the Mechanical advantage.(given g=9.8m/s2)

Dear Student, Mass of the bucket + water = M = 6 kg. Weight of the bucket + water = Mg = 6 kg×9.8 ms-2=58.8 N Mechanical Advantage (MA) = Force outputForce input=LoadEffort=58.870=0.84 Regards

Dear Student,
Mass of the bucket + water = M = 6 kg.
Weight of the bucket + water $= M g = 6 k g \times 9.8 m s^{- 2} = 58.8 N$
$M e c h a n i c a l A d v a n t a g e (M A) = \frac{F o r c e o u t p u t}{F o r c e i n p u t} = \frac{L o a d}{E f f o r t} = \frac{58.8}{70} = 0.84$
Regards