Question

A box of mass `4kg` rests upon an inclined plane This inclination is gradually incresed till the box starts sliding down the plane. At this stage slope of the plane is 1 in 3 Find coefficient of friction between the box and the plane What force applied to the box parallel to the plane will just make the box move up the plane ?

Answer 1

Correct Answer - `0.35 ; 26.13N`
Here, `m = 4kg sin theta = (1)/(3)`
`mu = ? F = `?
`mu = tan theta = (1)/(sqrt8) = 0.35`
As the box is to move up the plane force of friction is down the plane
`F =mg sin theta + f = mg sin theta + mu mg cos theta`
`=mg (sin theta + mu cos theta)`
`=4 xx9.8((1)/(3) + 0.35 sqrt8/(3)) = 26.13 N`
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