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A ladder AP of length 5m inclined to a vertical wall is slipping over a horizontal surface with velocity of 2m/s, when A is at a distance 3m from 0, the velocity of CM at this moment is: |
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Answer» Solution :From figure, `Y = sqrt(L^(2)-x^(2))` `(DY)/(DX) = x/sqrt(L^(2)-x^(2)) (dx)/(dt) =-(3 xx 2)/4 = -3/2 m//s` `V_(CM) = sqrt([1/2(dx)/(dt)]^(2) + [1/2(dy)/(dt)]^(2))` `=sqrt(1^(2) + (3/4)^(2)) = 1.25 m//s`
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