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Velocity Ratio

an effort of ...

Question

an effort of 6Kgf is applied on a machine through displacement of 90 cm,when a load of 100 kgf moves through a distance of 4 cm,Calculate i)velocity ratio ii)M.A iii)%of efficiency machine

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Solution

$V e l o c i t y r a t i o = \frac{d i s p l a c e m e n t b y t h e e f f o r t}{d i s p l a c e m e n t b y t h e l o a d} = \frac{90 c m}{4 c m} = 22.5 . M e c h a n i c a l a d v a n t a g e (M A) = \frac{L o a d}{E f f o r t} = \frac{100 K g f}{6 K g f} = 16.67 . E f f i c i e n c y = \frac{M A}{V e l o c i t y r a t i o} = \frac{100}{6} \div \frac{90}{4} = \frac{100}{6} \times \frac{4}{90} = \frac{40}{54} = \frac{20}{27} . E f f i c i e n c y % = \frac{20}{27} \times 100 = 74.7 % .$

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