Three force are applied to a square plate as shown in the figure. Find the magnitude, direction and point of application of the resultant force if this point is taken on the side `BC`.

Explanation

`F_(x)=F+sqrt2 F cos 45^(@) = 2F`
`F_(y)=F-sqrt2 F sin 45^(@) = 0`
Resultant force `vecF=2Fhati`
Resulant force of magnitude `2F` acting long the `+x`- direction.
Let `b` be the side of square and force `2F` is acting at distance `l` from `C`.
Taking torque of all forces about `C` and of resultant force about `C` and equating
`-Fxxb sin45^(@)-Fxx2b cos45^(@)+sqrt2 Fb=-2 F lsin45^(@)`
`l=(b)/(2)`
That is, resulant force is acting mid-point of `BC`.