A machine is operated by an effort of 80N acting downward and moving through a downward displacement of 015m. The load of mass 10kg is raised up by 10cm. Calculate:1) mechanical advantage2) velocity ratio 3)work input 4)useful work output 5)efficiency

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Solution

$M A = \frac{o u t p u t f o r c e}{i n p u t f o r c e} = \frac{l o a d}{e f f o r t} = \frac{100}{80} = 10 / 8 V e l o c i t y r a t i o = \frac{d i s \tan c e m o v e d b y e f f o r t}{d i s \tan c e m o v e d b y l o a d} = \frac{0.15 m}{10 c m} = 3 / 2 e f f i c i e n c y = \frac{M A}{V R} = \frac{10 / 8}{3 / 2} = 10 / 12 = 5 / 6 w o r k i n p u t = i n p u t f o r c e \times d i s \tan c e = 80 \times 0.15 = 12 j w o r k o u t p u t = o u t p u t f o r c e \times d i s \tan c e = 100 \times 10 / 100 = 10 j$

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A machine is operated by an effort of 80N acting downward and moving through a downward displacement of 015m. The load of mass 10kg is raised up by 10cm. Calculate:1) mechanical advantage2) velocity ratio 3)work input 4)useful work output 5)efficiency

MA = output forceinput force = loadeffort = 10080 = 10/8Velocity ratio = distance moved by effortdistance moved by load = 0.15 m10 cm =3/2efficiency = MAVR = 10/83/2 = 10/12 = 5/6work input = input force×distance = 80×0.15 = 12 jwork out put = output force ×distance = 100×10/100 = 10 j