The minimum force required to move a body up on an inclined plane is three times the minimum force required to prevent it from sliding down the plane.If the coefficient of friction between the body and the inclined plane is `1/(2sqrt3)` the angle of the inclined plane is

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `15^(@)`

Answer

C

Explanation

Minimum force required to move a body up a rough inclined plane is
`F_(1)=mg (sin theta + mu cos theta)`
Minimum force required to prevent the body form sliding down the rough inclined plane, is
`F_(2)=mu mg cos theta`
According to the question,
`F_(1)=3F_(2)`
`therefore mg (sin theta+mucos theta)=3(mu mg cos theta)`
`implies " "sin theta+mucos theta = 3 mu cos theta`
`implies " "sin theta=2mu cos theta`
`implies tan theta=2 mu = 2 xx (1)/(2sqrt(3))=(1)/(sqrt(3))=tan30^(@)`
` theta=30^(@)`