Equilibrium of rotating bodies - non-balancing vertical forces? I am slightly confused about this problem.
We have a $6$ kg non-uniform rod $MN$ which is pivoted about $M$. There is a force of $40$ N applied at $N$ at an angle to the rod.
The rod is said to be in equilibrium as a result of the $40$N force.
However, my question is that if we resolve vertically, we have the component of the $40$N force and the weight of the rod. These will not be equal, so why does the rod not fall downwards?
I know it has something to do with the pivot, but I can't see why the pivot will produce a vertical force.

 A: The rod is fixed to the pivot - hence it does not fall. The pivot here means something that keeps that end of the rod permanently in place, regardless of the forces exerted. It does this by exerting whatever force is necessary to counteract the other forces. If balancing the vertical forces causes a net force of, say, $20N$ downwards, then the pivot produces a force $20N$ upwards to counteract that.
As an illustration, consider Farcher's example of a trapdoor. If you pull one edge of the trapdoor, it opens. If you keep pulling it vertically, eventually the trapdoor is vertical. Now you can pull very hard and the trapdoor still won't move. That's because the other edge of the trapdoor is fixed to the floor, and that edge produces an almost-arbitrarily large force to balance the force you are pulling with.
A: The hinge at M provides a force of constraint. For equilibrium, the net torque about point M is zero (the force of constraint at M provides no torque about M); so you can calculate the center of mass distance along MN for equilibrium (the rod MN is not uniform).  The force at M is as required to counter the net force from gravity and the 40 N applied force.
This is similar to the constraint force provided by the support point of a lever; at equilibrium the net torque about the support point is zero and the force at the support balances the forces at each end of the lever.  The greater the total force at both ends of the lever, the greater the force of constraint at the support.
