A 5.00m long horizontal beam weighing 315N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of $53^o$ with the horizontal, and a 545N person is standing 1.50m from the wall. Find the force in the cable, $F_T$, and the force exerted on the beam by the wall, $R$, if the beam is in equilibrium.
Rework the example problem above with the axis of rotation passing through the center of mass of the beam. Verify that the answers do not change even though the axis is different.
In the original example problem, the fulcrum is placed at the "pin connection", which cancels out $R$. $F_T$ is solved using the torque equilibrium. However, with the fulcrum at the center of mass, there are no other given forces left, so there is only $R=-F_T$. While $l$ (5.00m, for R), $\theta$, and $d$ (1.50m) are given, you cannot (from what I can see) calculate anything relevant (a force or torque).
The forces were calculated to be
$F_T=403$N
$R=590$N
I was thinking maybe you could use $\tau_{net}=I\alpha$ -> $I=-\alpha$ to get a force but it doesn't really seem to be useful...
Hints: Your unknowns in this problem are the magnitude of force in the cable $F_T$, the magnitude of the reaction from the wall $R$, and the angle $\theta$ that $\mathbf R$ makes with the horizontal. So you need three equations. You can write equilibrium of forces in the vertical direction, in the horizontal direction, and the torque has to be zero with respect to any point on the bar. So you might choose the fulcrum, where the torque due to $\mathbf R$ is zero. So the torque due to the weight of the bar, and the weight of the person will cancel the torque due to the tension in the cable. Note that i believe that your $\alpha$ is the angular acceleration, and that is zero in the equilibrium.
Can you take if from here?