The force `F_(1)` that is necessary to move a body up an inclined plane is double the force `F_(2)` that is necessary to just prevent it form sliding down, then:

Options

A. `F_(2)=w sin ( theta-phi)sec phi`
B. `F_(1)=w sin (theta-phi)sec phi`
C. `tan phi= 3 tan theta`
D. `tan theta=3 tan phi` Where `phi=` angle of friction `theta=` angle of inclined plane `w=` weight of the body

Answer

A, D

Explanation

`F_(1)=mg sin theta + mu mg cos theta`
`F_(2)=mg sin theta-mu mg cos theta`
But `mg=w mu= tan phi`
`F_(1)=w(sin theta+(sin phi)/(cosphi)costheta)`
`rArrF_(1)=w sin (theta+phi)secphi`
`F_(2)=w (sin theta-(sin phi)/(cos phi)cos theta)`
`rArr wsin theta (theta-phi)sec phi`
Now `F_(1)=2F_(2)`
`mg sin theta+mu mg cos theta=2 (mg sin theta -mu mg cos theta)`
`sin theta+mu cos theta=2sin theta-2mucos theta)`
`rArr 3mu cos theta=sintheta`
`=tan theta=3 mu tan theta=3tan phi`