Question Video: Solving a Multi Step Problem Involving a Rigid Body in Equilibrium Under Parallel Forces Mathematics

Video Transcript

A nonuniform wooden board 𝐴𝐵, having a length of 16 meters, is resting horizontally on two supports at 𝐶 and 𝐷 such that 𝐴𝐶 is three meters and 𝐵𝐷 is four meters. If the maximum distance that a man whose weight is 639 newtons can move on the board from 𝐴 to 𝐵 without getting the board imbalanced is 14.2 meters and the maximum distance that the same man can move from 𝐵 to 𝐴 is 14.8 meters, find the weight 𝑤 of the board and the distance 𝑥 between its line of action and the point 𝐴.

There’s a lot of information given to us in this statement. The most important piece of information is that the board is resting, which means it is in equilibrium. And that tells us two conditions about the forces acting on the board. The conditions are that the sum of all the forces acting on the board is zero. And for any point, the sum of all the moments of force about that point is also zero. In this particular instance, our reference point will always be on the board, and all of the forces are acting perpendicular to the board. So we can express the moment of every force as the magnitude of the force times the distance from where the force is acting to the reference point. As we will discover, we can actually find everything that we’re looking for using only our condition for the moment of force, the definition of a moment of force, and some algebra.

Before we get there though, we need to draw a diagram to organize the information that we have. Actually, we’ll want to draw two diagrams, one corresponding to the man moving from 𝐴 to 𝐵 and one corresponding to the man moving from 𝐵 to 𝐴. We have our wooden board 𝐴𝐵, the two supports 𝐶 and 𝐷. And also, the length of the rod is 16 meters, the distance from 𝐴 to 𝐶 three meters, and the distance from 𝐷 to 𝐵 four meters. In terms of the forces, we know the weight 𝑤 is acting on the board somewhere between 𝐶 and 𝐷. We don’t know where because the board is nonuniform, but we know it is between 𝐶 and 𝐷 because the board is not tipping over.

The weight is one of the quantities that we’re looking for. The other quantity we’re looking for is the distance 𝑥 between where the weight is acting and the point 𝐴. It’s also worth including the distances from where the weight is acting to each support in our diagram. Because we don’t know 𝑥, we also don’t know the distance between where the weight is acting and the support 𝐶. So let’s just call this distance 𝑑.

There’re now a few other things that we can label in terms of 𝑑. For one thing, the distance from 𝐴 to the support 𝐶 is three meters and the distance from the support 𝐶 to where the weight is acting is 𝑑. So three plus 𝑑 is 𝑥. Furthermore, the total distance between support 𝐶 and support 𝐷 is 16 meters for the entire board minus four meters for 𝐷𝐵 minus three meters for 𝐴𝐶 or 16 minus seven or nine meters. But since the distance from support 𝐶 to where the weight is acting is lowercase 𝑑, the distance from where the weight is acting to the support 𝐷 must be nine meters minus lowercase 𝑑.

For the board to be in equilibrium, each support must also exert a reaction force pointing upward, which we’ll call 𝑅 sub 𝐶 and 𝑅 sub 𝐷. The only other force we have to consider is the 639-newton weight of the man. We’ll have this weight acting at two different points in the two different diagrams. In the first diagram, we’ll place the man 14.2 meters away from 𝐴, that is, as far as he can move from 𝐴 to 𝐵 without imbalancing the board. Since the board is 16 meters long, 14.2 meters away from 𝐴 is the same as 1.8 meters away from 𝐵. In the second diagram, we’ll place the man 14.8 meters away from 𝐵 since that’s as far as the man can move from 𝐵 to 𝐴 without tipping over the board.

As before, because the total length of the board is 16 meters, being 14.8 meters away from 𝐵 is the same thing as being 16 minus 14.8 or 1.2 meters away from 𝐴. Now our diagrams are finished. Although they’re quite full and there is a lot of information, we’ll see that by using what we already know and making a few clever choices, we can actually use each diagram to produce one simple equation. First, let’s use what we know.

We know that we’ve placed the man at the maximum distance possible without imbalancing the board. Let’s see what would happen if the man were moved a little bit farther, say in the second diagram a little bit closer to the end at 𝐴. If this happened, the board would start pivoting about the support at 𝐶 and, in fact, lift directly off of the support at 𝐷. If the board is lifting off of the support, it is obviously not exerting any pressure on the support. But if the board is not exerting any pressure on the support, then the support is not exerting any reaction force on the board. But because there is no pressure when the board lifts away from 𝐷, there is also no pressure when the board is exactly balanced and the man is 1.2 meters away from 𝐴.

So in our second diagram where the board is exactly balanced about the support 𝐶, 𝑅𝐷 is zero. Similarly, in the first diagram where the board is exactly balanced about the support at 𝐷 with the man on one side and the weight on the other, there is no pressure on the support at 𝐶 and our 𝐶 is zero. We have now eliminated one reaction force from each of our diagrams. Let’s now take a moment to think about our conditions for equilibrium.

In our expression for a moment of force, if the force is acting at the reference point, then the distance from the force to the reference point is zero and the moment of the force about that reference point is zero. So, in our first diagram, if we choose the reference point where the support 𝐷 meets the board, then the moment of 𝑅𝐷 about that reference point will be zero. Similarly, in the second diagram, if we choose the reference point to be where the support at 𝐶 meets the board, then the moment of 𝑅𝐶 about that reference point will be zero. Furthermore, if the magnitude of a force is zero, then the moment of that force about any reference point is also zero.

So, in the first diagram, the reaction at 𝐶 is zero. And in the second diagram, the reaction at 𝐷 is zero. And thus both of these reactions provide zero moment about their respective reference points. This leaves us in both diagrams with the weight of the board and the weight of the man as the only two forces we need to consider when applying our condition for the net moment of force. Also note that, in both diagrams, the weight of the board and the weight of the man are acting in the same direction, but on opposite sides of the reference point. So their moments have opposite signs.

All right, let’s now calculate the net moment for each diagram. In the first diagram, the weight 𝑤 is acting nine meters minus lowercase 𝑑 away from the reference point. So its moment is 𝑤 times nine minus lowercase 𝑑, which we’ve arbitrarily chosen to be positive. This means that the moment of 639 newtons, the weight of the man, will be negative. Well, we need to know the distance between where this force is acting and the reference point.

Since the man is 1.8 meters away from 𝐵 and 𝐵 is four meters away from the reference point, the man must be 2.2 meters, four minus 1.8, away from the reference point. So the net moment in our first diagram about the point 𝐷 is 𝑤 times nine minus 𝑑 minus 639 times 2.2, which from our conditions for equilibrium must be equal to zero.

In our second diagram, the reference point is where the support 𝐶 meets the board. So the weight of the board is a distance 𝑑 away from this reference point. This time, the weight of the man is 1.2 meters away from 𝐴, and 𝐴 is three meters away from the reference point. So the man is 1.8 meters away from the reference point. Combining all this, we have a net moment of 𝑤 times 𝑑 minus 639 times 1.8, which again equals zero.

Note that we have again taken care that the moments of the two forces have different signs when they’re acting in the same direction but on opposite sides of the reference point. And again, we have arbitrarily chosen our sign convention so that the weight of the board has a positive moment. As we mentioned before, we’re allowed to do this as long as we choose consistently with any particular diagram.

Now we have two equations with two unknowns. And all that’s left to do is use some algebra to solve. Let’s now rearrange both of these equations into a more useful form. In the first equation, we’ll distribute 𝑤 over nine minus 𝑑, and we’ll also add 639 times 2.2 to both sides. And in the second equation, we’ll add 639 times 1.8 to both sides. We have nine 𝑤 minus 𝑤𝑑 equals 639 times 2.2 and 𝑤𝑑 equals 639 times 1.8. In this form, we see that we can eliminate 𝑑, more precisely 𝑤𝑑, from our first equation by adding our second equation.

Let’s clear some space to do this calculation. We’ll remove the original equations we wrote and keep these second forms because they convey the same information but in a more useful way. When we add these equations together, on the left-hand side, 𝑤𝑑 plus negative 𝑤𝑑 is zero. So we’re just left with nine 𝑤. On the right-hand side, we have 639 times 2.2 plus 639 times 1.8, which when we plug into a calculator we get 2556.

To isolate 𝑤, all we need to do is divide 2556 by nine. When we do this, we find that 𝑤, the weight of the board that we’re looking for, is 284 newtons. Now we use either of our two equations substituting in for 𝑤 to find 𝐷. And then from 𝐷, we can find 𝑥, which is the other quantity that we’re looking for. Let’s use our second equation 𝑤𝑑 equals 639 times 1.8, substituting in 284 for 𝑤. We isolate 𝑑 by dividing 639 times 1.8 by 284. This gives us 𝑑 is exactly 4.05. And adding three to find 𝑥, 𝑥 is 7.05 meters.

So even though we started with a lot of information and a rather complicated diagram, we were able to extract from that two relatively simple equations and then from that solve for the weight of the board and also the location where that weight is acting.