Pushing force making an angle o to the horizontal is applied on the block of weight W placed on the horizontal table. If o is the angle of friction, the magnitude of the force required to move the body is equal to :

Options

A. `(Wcos phi)/(cos(theta-phi))`
B. `(Wsin phi)/(sin(theta-phi))`
C. `W sin phi//cos(theta-phi)`
D. `W tan theta//sin(theta-phi)`

Answer

c

Explanation

`F_(r)`= force of friction. Then `tan phi =(F_(r))/(R)`

or `F_(r) = R tan phi` where phi is angle of friction. The block will be just at the point of motion when
`F cos theta =F_(r) = R tanphi`and `F sin theta+ R = W`
Now `R=W-F sin theta`
Or `F cos theta = (W-F sin theta) tanphi`
`F cos theta+F sin theta (sin theta)/(cos phi)=(W sin phi)/(cos phi)`
or` F cos theta cos phi +F sin theta sin phi = W sin phi`
`F(cos theta cos phi + sin theta sin phi) = W sin phi`
or`F=(Wsin theta phi)/(cos (theta-phi))`