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Net horizontal force, H = f + f

2cos 45 = 2f
Net vertical force, V = f - f

2 sin45 = 0
Hence the net force is horizontal, ie, parallel to AC.
Magnitude of resultant force, R = 2f
Let this resultant force be applied at a point x distant from C on the side BC.
Then the torque due to this force at any point must equal the sum of torques of the forces shown.
Taking torques about the point of intersection of the diagonals,
2f xsin45 = f(L/

2) + f(L/

2) - (f

2cos 45)((L/

2)
x = L/2
where L is the length of the side of the square.
Hence the resultant is applied at the midpoint of the side BC

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