1.

Three blocks A,B and C of mass m,m/2 and m of different densilties and dimensions are placed over each other as shown in the figure. The cofficients of friction are shown. Blocks placed in a vertical line are mae to over towards right with same velcity at the same instant. Find the time (in sec) taken by the upper block A to topple from the middle blcok B. Assume that blocks B and C don't stop sliding before A topples from B. (given L=36m,mu=0.4 and g=10 m//s^(2))

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Solution :Since `(mu)/4lt(mu)/4ltmu`, the upper block will move faster than the MIDDLE block and HENCE force of friction on upper block is towards left
`:.f=(mu)/4mg`
`:.a_(A)=(mug)/4` leftward

`f^(')=(mu)/2(m+m/2)g=3/4mumg`
`:.a_(B)=mug` (leftward) ltbgt `:.a_(A//B)=3/4 mug` (rightward)
For sliding down the block `B, A` has to move a distance `(3L)/8`
relative to block `B`.
`(3L)/8= 1/2 3/4 mu"gt"^(2)`
`:.t=sqrt(l/(mug))=3` SEC


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