A nonuniform rod 𝐴𝐵 having a weight of 40 newtons and a length of 80 centimeters is suspended vertically from its midpoint by a light string, and it becomes in equilibrium in a horizontal position when a weight of magnitude 29 newtons is suspended from its end 𝐴. Determine the distance 𝑥 between the point at which the weight of the rod is acting and end 𝐴. After removing the weight at 𝐴, determine the magnitude of the vertical force that would be needed to keep the rod in equilibrium in a horizontal position when it acts at end 𝐵.
For both parts of this question, finding the distance 𝑥 and the magnitude of the vertical force acting at 𝐵 when the weight is removed, we were told that the rod will be in equilibrium, which means that two conditions are met. First of all, the sum of all the forces acting on the rod is zero. In other words, there is no net force. Secondly, the sum of all the moments of force about any point is also zero. So there is no net moment acting anywhere on the rod. If the direction of a force is perpendicular to the line connecting where that force is acting and some reference point, then the moment of that force about the reference point is equal to the magnitude of the force times the distance between where the force is acting and the reference point.
We can now use these conditions and the information we’re given about the forces and distances to solve for what we’re looking for. Let’s start by drawing a diagram to organize the information we have about what forces are present and where they are acting. Here is our nonuniform rod suspended from its midpoint by a light string. We’ve drawn the rod as perfectly horizontal because we were told in the statement that it is horizontal in equilibrium.
Let’s now add some forces to the diagram. In order to maintain this equilibrium, we’re told that there is a weight with a magnitude of 29 newtons that is suspended from the end 𝐴. We’re also told that the weight of the rod is 40 newtons, but because the rod is nonuniform, we don’t know where this weight is acting. We do, however, know that this weight must be acting closer to 𝐵 than to 𝐴 because otherwise all of the forces would be on one side of the midpoint and the rod would be unbalanced.
Finally, there’s one more force not explicitly mentioned in the statement. This is the tension in the string which is actually holding the rod up against the downward pull of the rod’s own weight and the additional weight at 𝐴. Since we don’t have a value for this tension force, we’ll just call it 𝑇. To complete this diagram, we need to include the distances that we’re given. We know that the total length of the rod is 80 centimeters, so the distance from the midpoint where the string is attached to either end is 40 centimeters. And the distance from 𝐴 to where the weight of the nonuniform rod is acting is defined as the distance 𝑥.
Okay, now all we need to do is apply our conditions for equilibrium to solve for what we’re looking for. A sensible approach would be to use our condition for the forces to find the unknown tension in the string and then use our condition for the moments of force to determine 𝑥. If we take this approach, it turns out that it won’t matter what we choose for our reference point when we calculate those moments of force. However, in order to illustrate the importance of reference points when it comes to calculating moments of force, we will actually cleverly choose our reference point so we only need to use this second condition and never need to calculate the value of the unknown tension 𝑇.
In order to remove the tension from the calculations, note that the moment of force is defined as the magnitude of the force times the distance to the reference point. So if the force is acting exactly at the reference point, the distance to the reference point is zero. And so the moment of that force about the reference point is also zero. So we’ll take our reference point for calculating moments as the center of the rod. Since the tension is acting at the midpoint of the rod, the moment of the tension about this point is zero. So all we need to do is balance the moments of the weight of the rod itself and the moment of the additional weight attached at 𝐴.
To set up the equation, note that 29 newtons and 40 newtons are both acting in the same direction, but they’re acting in opposite orientations relative to our reference point. We can see this by imagining a circle drawn around the reference point. A downward force to the right of the reference point is going clockwise around the circle, while a downward force to the left of the reference point is going counterclockwise around the circle. What this means is that the moments from each of these forces about the reference point have opposite signs.
Let’s arbitrarily take the moment of the force acting at 𝐴 to be positive so the weight of the rod itself is providing a negative moment. The distance from the force at 𝐴 to the midpoint is 40 centimeters, so its moment is 29 times 40. The weight of the rod has a magnitude of 40 newtons, and it is acting this distance away from the reference point. Although we don’t know this distance directly, we know that this distance from 40 newtons to the midpoint plus 40 centimeters from the midpoint to 𝐴 is exactly the distance 𝑥. So this distance is actually 𝑥 minus 40.
So the moment of the weight of the rod about its midpoint is negative 40 times 𝑥 minus 40, and the total moment is zero. Now we just need to solve this equation for 𝑥. Both terms have a factor of 40, so let’s divide both sides by 40. We have 29 minus 𝑥 minus 40 equals zero because zero divided by 40 is still zero. Adding 𝑥 minus 40 to both sides, we have that 29 is equal to 𝑥 minus 40, which we’ll replace in our diagram because we’re going to need it later. Finally, adding 40 to 𝑥 minus 40 gives us 𝑥 and adding 40 to 29 gives us 69. So 𝑥, the distance from where the weight is acting to the end 𝐴, is 69 centimeters. And that’s half our answer.
Now we need to consider a different situation. The weight at 𝐴 has been removed. And now there’s a vertical force instead acting at 𝐵 to keep the rod in equilibrium. Let’s modify our diagram to reflect this change. We’ve removed the 29-newton weight acting at 𝐴 and added a vertical force with unknown magnitude 𝐹 acting at 𝐵. Since the rod is still in equilibrium in its horizontal position, we know that our previous conditions still hold. Also just like before, we only need the condition for the moments of force as long as we take the reference point to be the center of the rod so the effects of the unknown tension disappear.
Let’s clear off our previous calculations to make room for the new one. There are now two forces we need to consider, the 40-newton weight acting 29 centimeters from the midpoint of the rod and the vertical force 𝐹 acting 40 centimeters from the midpoint of the rod. If we keep our previous choice for the sign of the moments, the 40-newton force is still clockwise about the reference point. So its moment is negative 40 times 29. However, the vertical force at 𝐵, although it is also to the right of the reference point, is pointing upward not downward. So it is pointing in the counterclockwise orientation relative to the reference point, so its moment is positive.
So we have positive 𝐹 times 40, the distance from 𝐵 to the midpoint of the rod, as the moment for the force 𝐹. And as before, this gives us an expression, negative 40 times 29 plus 𝐹 times 40, that must evaluate to zero. There’s a factor of 40 in both terms on the left-hand side, so we divide both sides by 40 relying on the fact that zero divided by 40 is still zero. This gives us negative 29 plus 𝐹 equals zero. And adding 29 to both sides, we find that 𝐹, the magnitude of the force acting at 𝐵 that keeps the rod in equilibrium, is exactly 29 newtons.
It’s worth stressing again that the reason we were able to find these answers without reference to the tension in the string is because we cleverly chose our reference point so that the tension played no role in our calculation of the net moment. It’s also worth mentioning that this is only useful because we weren’t looking to find the tension in the string. If we had been looking to find the tension, we could’ve always found it by using our condition for the net force.