Non-uniform beam weight distribution

[greg_rack]

Homework Statement

A non-uniform beam PQ of length 5.0 m and weight X rests on a pivot placed 3.0 m from end P. It is kept in equilibrium in a horizontal position by an upward force of magnitude 0.60X acting at end P.
The normal contact force at the pivot is 800 N.
What is the weight of the beam and how far is the center of gravity of the beam from the pivot?

Relevant Equations

W=mg
Ftot=0
Mtot=0

Since the "non-uniform" hypothesis I got confused about how to solve this problem... in which way could I find its center of mass, if its mass is not distributed uniformly?
Another question I have with regards to this static situation is: is the beam placed horizontally? If not, how could I know the angle formed by the normal contact force with the beam exerted by the pivot?

 

[Chestermiller]

Homework Statement:: A non-uniform beam PQ of length 5.0 m and weight X rests on a pivot placed 3.0 m from end P. It is kept in equilibrium in a horizontal position by an upward force of magnitude 0.60X acting at end P.
The normal contact force at the pivot is 800 N.
What is the weight of the beam and how far is the center of gravity of the beam from the pivot?
Relevant Equations:: W=mg
Ftot=0
Mtot=0

Since the "non-uniform" hypothesis I got confused about how to solve this problem... in which way could I find its center of mass, if its mass is not distributed uniformly?
Another question I have with regards to this static situation is: is the beam placed horizontally? If not, how could I know the angle formed by the normal contact force with the beam exerted by the pivot?

Assume the beam is horizontal. You are only looking for the location of its center of mass, so you don't need to know the mass distribution.

 

[greg_rack]

Assume the beam is horizontal. You are only looking for the location of its center of mass, so you don't need to know the mass distribution.

I'm actually feeling quite dumb right now, I've solved it