Equation for gravitational torque

[alem]

Homework Statement

A thin rod (uniform density and thickness)
with mass M and length L attached to floor at a fixed location by a frictionless hinge.

while balanced vertically the gravitational torque acting on the rod is
a.) zero
b.) 1/2 MgL
c.)1/3 MgL
d.)1/4 MgL
e.)1/6 MgL

Homework Equations

I know that torque is F x D(or L)
F=Ma
and a is g in this case,
so T= MgL

The Attempt at a Solution

as stated above I got to T=MgL
but I don't understand where the fractions are coming from? can someone please explain this?

 

[JHamm]

You forgot something when computing the cross product, there's a trig function in there somewhere ;)