Three blocks of mass m1 = 2 kg ,m2 = 4 kg, and m3 = 6 kg are connecyed by light strings on a frictionless inclind plane of 60 degree . A force 'F' of 120 N is applied upwards along the incline to the uppermost block, causing an upward movement of the block. Calculate the acceleration of the blocks, and tension in the strings. Take g = 10 m/s .

Asked by Mohit RaAz | 22nd Aug, 2015, 08:13: PM

Expert Answer:

 
begin mathsize 12px style Consider space three space block space of space mass comma space straight m subscript 1 space equals space 2 space kg comma space straight m subscript 2 space equals space 4 space kg space and space straight m subscript 3 space equals space 6 space kg space are space connecyed space by space light space strings space on space straight a space frictionless space inclind space plane. An space inclined space plane space makes space an space angle space 60 degree space with space horizontal. Let space straight F space be space the space force space of space 120 space straight N space applied space upwards space along space the space incline space to space the space uppermost space block. where space straight g space equals space 10 space straight m divided by straight s squared The space acceleration space due space to space gravity space along space the space inclined space equals space straight g space sin space straight theta space equals space 9 space sin space 60 space Applying space Newton apostrophe straight s space Law comma straight F space minus space mg space sin space 60 space equals space ma 120 space minus space 12 space cross times space 10 space cross times space fraction numerator square root of 3 over denominator 2 end fraction space equals space 12 space cross times space straight a space space space space space space space space space left parenthesis where space straight m space equals space straight m subscript 1 space plus space straight m subscript 2 space plus space straight m subscript 3 space and space straight a space is space the space acceration space of space the space block right parenthesis Hence comma space we space get space space straight a equals 1.34 space straight m divided by straight s squared Tension space in space the space string space straight T space is space given space as comma straight T subscript 1 space equals space straight m subscript 1 straight g space sin space straight theta space space equals space 2 space cross times space 10 space cross times space sin space 60 straight T subscript 1 space equals space 2 space cross times space 10 space cross times space fraction numerator square root of 3 over denominator 2 end fraction space equals space 17.32 space straight N  straight T subscript 2 space equals space straight m subscript 2 straight g space sin space straight theta space equals space 4 space cross times space 10 space cross times space sin space 60 straight T subscript 2 space equals space 4 space cross times space 10 space fraction numerator square root of 3 over denominator 2 end fraction space equals space 34.64 space straight N   end style

Answered by Priyanka Kumbhar | 24th Aug, 2015, 09:35: AM

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