The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is `0.25` , the angle of inclination of the plane is

Options

A. `37^@`
B. `45^@`
C. `30^@`
D. `53^@`

Answer

A

Explanation

Retardation in upward motion `=g(sintheta+mucostheta)`
Force required just to move up `F_(up)=mg(sintheta+mucostheta)`
Similarly for downward motion `a=g(sintheta+mucostheta)`
Force required just to prevent the body sliding down
`F_(db)=mg(sintheta-mucostheta)`
According to problem `F_(up)=2F_(dn)`
`impliesmg(sintheta+mucostheta)=2mg(sintheta-mucostheta)`
`impliessintheta+mucostheta=2sintheta-2mucostheta`
`implies3mucostheta=sinthetaimpliestantheta=3mu`
`impliestheta=tan^-1(3mu)=tan^-193xx0.25)=tan^-1(0.75)`