A uniform door weighs 51.0 N and is 1.0 m wide and 2.5 m high. What is the magnitude of the torque due to the?

Question

A uniform door weighs 51.0 N and is 1.0 m wide and 2.5 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?

so far i've done: Ux=(51*1)/(2*25) =10.2N
Newtons 3rd Law=Dx=10.2N and Dy=51N
51N/10.2N =5 Nm but that's wrong. Anyone care to show me how to do this problem? Thanks!

Answer 1

Assuming the door is upright, the weight of the door will act from the centre of mass along a line whose perpendicular distance to the pivot is 0.5m.
Torque=Force * Perpendicular distance
T=51*0.5=25.5Nm

Answer 2

T = (51.0 N)(0.5 m) = 25.5 N-m