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

Explanation

For coplanar forces, about any point in the same plane, `sumvecr_ixxvecF_i=vecrxxvecF_(n et)`
(where `vecF_(n et)=sum vecF_i` = resultant force) or, `vecN_(n et)=vecrxxvecF_(n et)`
Thus length of the arm, `l=(N_(n et))/(F_(n et))`
Here obviously `|vecF_(n et)|=2F` and it is directed toward right along AC. Take the origin at C. Then about C,
`vecN=(sqrt2aF+a/sqrt2F-sqrt2aF)` directed normally into the plane of figure.
Here `a=` side of the square.)
Thus `vecN=F(a)/(sqrt2)` directed into the plane of the figure.
Hence `l=(F(a//sqrt2))/(2F)=(a)/(2sqrt2)=a/2sin45^@`
Thus the point of application of force is at the mid point of the side BC.