Static equilibrium: Placing a fulcrum

[archaic]

Homework Statement

A 0.140-kg, 45.4-cm-long uniform bar has a small 0.050-kg mass glued to its left end and a small 0.100-kg mass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.
How far from the left end should the fulcrum be placed?

Relevant Equations

##\sum\tau=0##

I don't understand the question; they're telling me that the fulcrum is just under the center of gravity of the bar, but that I also need to find its position.
In any case, I suppose that the fulcrum's position is ##x##.

Let the weight at the left end be ##w_1##, at the right end be ##w_2##, and of the center of gravity be ##w##.
Since ##w_2>w_1##, I'm going to suppose that ##x>\frac l2##.
$$w(x-\frac l2)+w_1x=w_2(l-x)\\x=l\frac{w_2+\frac w2}{w+w_1+w_2}=45.4\times\frac{0.100+\frac{0.140}{2}}{0.140+0.100+0.050}$$

 

[archaic]

Ok, this is correct. I have first mindlessly put ##x## randomly less than ##l/2## and worked that way.

 

[archaic]

Generally, do engineers balancing such systems check from which side of the center of gravity the sum of the forces' component contributing to rotation is greater before setting up the equation? Or there is a systematic way?

 

[Cutter Ketch]

If it isn’t moving the sum of the torques about any point must be zero. Pick a point, pick a sign convention for the torques, and the answer will come out right no matter what point you pick. Engineers don’t generally try to guess the right point. Instead they generally pick a point to make the math easy. For example, if there’s a force you don’t know use it’s location as the center of rotation and it’s torque is zero no matter what the magnitude. It drops out of the calculation. Similarly pick a place where several forces act to get them all out. Perhaps most the forces are perpendicular but one is at an unknown angle and you don’t want to have to figure out the components. Choose that as the origin. Etc.