# Lesson Video: Equilibrium of a Rigid Body under Parallel Forces Mathematics

In this video, we will learn how to solve problems about the equilibrium of a body under the effect of parallel coplanar forces.

18:00

### Video Transcript

In this lesson, we’ll learn how to solve problems that involve the equilibrium of a body under the effect of parallel coplanar forces. Let’s begin by recalling some of the key facts and terms that we use when working with forces. Informally, we consider that when a body is in equilibrium, it’s kind of balanced. And so, formally, we say that for a body to be in equilibrium, the sum of the forces acting on that body must be equal to zero.

Now, since force can have direction as well as magnitude, we might also say that the vector sum of the forces is zero. Alternatively, if we think about a force acting in two directions, we can define a direction for the force in terms of the 𝑥- and 𝑦-direction. And we can say that the sum of the forces in each direction is zero. The sum of 𝐹 sub 𝑥 is zero, and the sum of 𝐹 sub 𝑦 is zero.

Similarly, for a body to be in equilibrium, we say that the sum of the moments must be equal to zero, where the moment is the turning effect of the force. The formula we use to calculate a moment is 𝐹𝑑. Now, 𝐹 here is the force that acts at a point, whereas 𝑑 is the perpendicular distance from the line of action of the force to the point about which the object will turn. When thinking about the equilibrium of a rigid body, such as a rod, under a system of forces, we often need to combine both of these ideas to help us find missing information. Let’s see what that might look like.

A uniform rod having a weight of 35 newtons is resting horizontally on two supports 𝐴 and 𝐵 at its ends, where the distance between the supports is 48 centimeters. If a weight of magnitude 24 newtons is suspended at a point that is 38 centimeters away from 𝐴, determine the reactions of the two supports 𝑅 sub 𝐴 and 𝑅 sub 𝐵.

When presented with a question like this, before doing anything, we begin by drawing a free body diagram. This is a very simple diagram that highlights the key forces that are acting on our body. Here is our 48-centimeter rod resting on supports 𝐴 and 𝐵 at its ends.

Next, we’re told that the rod is uniform. And this means that the downwards force of its weight, that’s 35 newtons here, must act at a point exactly halfway along the rod itself, so 24 centimeters then from either end. We then have another downwards force. There is a weight of magnitude 24 newtons suspended at a point 38 centimeters away from 𝐴.

So this is everything that we’ve been given in the question, but we’re not quite finished with our diagram. Newton’s third law of motion, which is often informally stated as every action has an equal and opposite reaction, tells us that since the rod is exerting a downwards force on both supports 𝐴 and 𝐵, there must be a reaction force of the supports on the rod itself. These reaction forces are normal, that’s perpendicular, to the rod. And we’re going to call them 𝑅 sub 𝐴 and 𝑅 sub 𝐵. And so now we have all of our forces, what do we do next?

Well, we’re told the rod is resting, so we’re going to make an assumption. And that assumption is that the rod itself is in equilibrium. For a body to be in equilibrium, there are two important criteria. The first is that the sum of all the forces acting on that body must be equal to zero. And the second is that the sum of all the moments must also be equal to zero, where the moment, which is essentially the turning effect of the force, is calculated by multiplying 𝐹 by 𝑑. 𝐹 here is the force that acts at a point, and 𝑑 is the perpendicular distance from the line of action of this force to the point about which the object will turn.

So let’s begin by forming an equation involving the sum of the forces. In this case, forces are only acting in the vertical direction. So we’re going to define a positive direction here. We’re going to define upwards to be positive so that 𝑅 sub 𝐴 and 𝑅 sub 𝐵 are acting in the positive direction. And then 35 and 24 are acting in the negative direction. We can therefore say that the sum of the forces acting in this direction must be 𝑅 sub 𝐴 plus 𝑅 sub 𝐵 minus 35 minus 24. And of course we know that the sum of these forces must be equal to zero for the body to be in equilibrium. Negative 35 minus 24 is negative 59. So we’re going to add 59 to both sides of this equation. And we find that 𝑅 sub 𝐴 plus 𝑅 sub 𝐵 is equal to 59.

And there’s not a lot else we can do with the sum of the forces. And so we move on to the second piece of information. That is, the sum of the moments of our forces must also be equal to zero. And so we choose a point about which to take moments. Now, it really doesn’t matter the point about which we choose to take the moments as long as we’re very careful to define a direction. In this case, we’re going to take moments about one of the endpoints of our rod.

Let’s take moments about 𝐴 since we’ve defined all of the distances from this point. And we’re going to define the counterclockwise direction to be positive here. And of course we define a moment as force times the perpendicular distance from the line of action of the force to the point about which the object will turn.

It’s worth noting that we generally calculate moments in terms of newton meters. And so our forces are in newtons and our distance is in meters. However, in this case, we’re working with newtons and centimeters. And it’s absolutely fine to work in newton centimeters as long as we’re consistent. So let’s begin by calculating the moment of this 35-newton force. It’s trying to turn the body in a clockwise direction, so the moment is going to be negative. And it’s acting at a point 24 centimeters away from 𝐴. So the moment is going to be negative 35 times 24.

Working from left to right, we now see we have to deal with this 24-newton force. Once again, this is trying to move the body in a clockwise direction. So its moment is going to be negative. Force times distance here is 24 times 38. Then there’s one more moment we need to consider. And that’s the moment of the reaction force at 𝐵. This time, this is acting in a counterclockwise direction, so the moment is going to be positive. It’s 𝑅 sub 𝐵 times the distance from 𝐴, which is 48. So that’s the sum of our moments. And of course we know those are equal to zero. Negative 35 times 24 minus 24 times 38 is negative 1752. So our equation simplifies as shown.

We can now solve this equation for 𝑅 sub 𝐵 by adding 1752 to both sides. So 48𝑅 sub 𝐵 is 1752. Then, finally, we divide through by 48. 1752 divided by 48 is 36.5. And so we can say that 𝑅 sub 𝐵 is 36.5 newtons. We still need to calculate the value of 𝑅 sub 𝐴. So we’re going to clear some space and go back to this equation here. 𝑅 sub 𝐴 plus 𝑅 sub 𝐵 is 59. We can now replace 𝑅 sub 𝐵 with 36.5. And so this equation becomes 𝑅 sub 𝐴 plus 36.5 equals 59. Let’s solve for 𝑅 sub 𝐴 by subtracting 36.5 from both sides. So 𝑅 sub 𝐴 is 59 minus 36.5, which is 22.5. So we’ve calculated the reactions of the two supports. 𝑅 sub 𝐴 is 22.5 newtons, and 𝑅 sub 𝐵 is 36.5 newtons.

So we saw that by considering the forces and the moments acting on the body simultaneously, we could create a system of linear equations which we could solve to find missing forces. We can also use this very same process to find other missing information, such as the length of the rod. Let’s have a look at an example of this.

In the figure, forces having magnitudes of 61, 43, 100, and 𝐹 newtons are acting on the light rod, and the rod is in equilibrium horizontally. Determine the length of the line segment 𝐷𝐴 and the magnitude of 𝐹.

The key to answering this question is to spot that the rod is in equilibrium. So what does that mean? Well, it means two things. Firstly, the sum of all the forces acting on the rod is equal to zero. In this case, we’ll consider the sum of the forces acting in a vertical direction. Secondly, it also means that the sum of the moments of our forces is also equal to zero, where the moment is calculated by multiplying the force acting at a point by the perpendicular distance from the line of action of this force to the point about which the object will turn.

So let’s begin by considering the sum of the forces acting on our diagram. Let’s define the positive direction to be upwards so that 𝐹 is acting in the positive direction, whilst 61, 43, and 100 are acting in the negative direction. We can therefore say that the sum of our forces must be 𝐹 minus 61 minus 43 minus 100. And of course we know that this sum is equal to zero.

Now, in fact, the expression on the left-hand side simplifies to 𝐹 minus 204. So 𝐹 minus 204 is equal to zero. We’ll solve for 𝐹 by adding 204 to both sides. So 𝐹 is equal to 204 or 204 newtons. So we’ve now calculated the value of 𝐹 and we’ve done everything we can with the sum of our forces. So we move on to the second criteria. The sum of the moments of our forces is equal to zero.

Now, we’ve been defined a positive direction here. That’s the counterclockwise direction. And we know that a moment is calculated by multiplying the force by the perpendicular distance of this force from the point about which the object is trying to rotate. We have a few distances on here, but there is one distance that we’re missing.

Let’s define the distance between 𝐴 and 𝐷, so the length of the line segment 𝐷𝐴, which coincidently we’re trying to find, to be 𝑥 centimeters. And then once we have that information, we pick a point about which to calculate our moments. Now, we can pick any point on the rod itself. We’re going to pick 𝐷 here. Now, the reason we’re going to pick 𝐷 is because the force 𝐹 is acting at this point. And had we not calculated 𝐹 yet, we still would’ve been able to calculate the moments about this point since the moment of 𝐹 would’ve been zero.

It’s also worth noting that we generally choose to work in newton meters when calculating moments. But actually, our dimensions are in centimeters. So we’re going to be working in newton centimeters throughout. And that’s absolutely fine as long as we’re consistent. So let’s find the moment of our 61-newton force. This force is trying to turn the object in a counterclockwise direction. So its moment will be positive. It’s 100 centimeters away from 𝐷, so the moment is 61 times 100.

Moving from left to right, we’re now going to deal with the force at 𝐶. Once again, this is trying to move the object in a counterclockwise direction. So its moment is positive. But now it’s 43 times 50. As we said, the force 𝐹 is zero centimeters away from 𝐷. So we don’t need to worry about that moment. And instead we move on to the force at 𝐴.

Now, this force is trying to move the object in a clockwise direction. So its moment is actually going to be positive. And it’s 100 times the distance from 𝐷, which is 𝑥. We know of course that the sum of these moments is zero. So we can form an equation in 𝑥. This equation simplifies to 8250 minus 100𝑥 equals zero. Then we add 100𝑥 to both sides and finally divide through by 100. So 𝑥 is 82.5. And we see that the length of line segment 𝐷𝐴 is 82.5 centimeters.

In our final example, we’ll see how we can use the information about an object in equilibrium and a little bit of intuition to model problems in which a rod is on the point of rotating.

The length of a rod 𝐴𝐵 is 111 centimeters, and its weight is 95 newtons, which is acting at its midpoint. The rod is resting horizontally on two supports, where one of them is at end 𝐴 and the other is at a point 𝐶, which is 30 centimeters away from 𝐵. A weight of 71 newtons is suspended from the rod at a point that is nine centimeters away from 𝐵. Find the magnitude of weight 𝑤 that should be suspended from end 𝐵 so that the rod is about to rotate, and determine the value of the pressure 𝑃 exerted on 𝐶 in that situation.

Let’s begin by drawing a free body diagram showing this scenario. Here is our rod. Now, the downwards force of its weight acts at its midpoint. So that’s 55.5 centimeters from either end. Now, we also have this 71-newton force. Now, that’s acting downwards at a point nine centimeters away from 𝐵. We want to add a weight 𝑤 at point 𝐵. So we add that to the diagram and two further forces. Those are the reaction force of the support on the rod. We’ll call them 𝑅 sub 𝐴 and 𝑅 sub 𝐵, respectively.

Now, the rod is about to rotate, so it’s essentially in limiting equilibrium. So we can say two things. Firstly, the sum of all the forces acting on the body is zero. And secondly, the sum of the moments is also equal to zero. So we’ll begin by considering the sum of the forces. We’re going to take the direction in which the reaction forces are acting as being positive.

And so this means that the 95-newton force, the 71-newton, and the weight force 𝑤 must be acting in the opposite direction. And so we can therefore say that the sum of the forces, which we know is equal to zero, is 𝑅 sub 𝐴 plus 𝑅 sub 𝐶 minus 95 minus 71 minus 𝑤 equals zero. And if we add 95 and 71 to both sides of this equation, we get 𝑅 sub 𝐴 plus 𝑅 sub 𝐶 minus 𝑤 equals 166.

There’s not a lot more we can do with this. And so we move on to the next bit of information. And this is the sum of the moments of our forces is also equal to zero. Let’s clear some space and find a point about which we want to take moments.

Now, since we have this unknown force 𝑤 at 𝐵, let’s take moments about 𝐵. And of course we define a direction to be positive. This time, let’s choose counterclockwise. We begin by thinking about the reaction force at 𝐴. We know that this is trying to move the body in a clockwise direction. And so its moment is going to be negative. It’s negative 𝑅 sub 𝐴 times 111.

The moment of the weight force is acting in a counterclockwise direction, so it’s positive. It’s 95 times 55.5. We then have negative 𝑅 sub 𝐶 times 30. That’s the moment of the reaction force at 𝐶. And then we have 71 times nine. That’s the moment of this 71-newton force. The sum of these moments, which we’re of course measuring in newton centimeters here, is equal to zero. And this simplifies as shown.

Now, there’s one bit of information that we haven’t yet used. And that is that the rod is on the point of rotating. Now, since we’re adding a weight at 𝐵, we know the object is going to be rotating about 𝐶. And so this means that, at this exact moment, there must be no downwards force of the rod on point 𝐴. And so, similarly, the reaction force at 𝐴 must also be equal to zero. So we can rewrite our equation by letting 𝑅 sub 𝐴 be equal to zero. And we’ll solve by adding 30𝑅 sub 𝐶 to both sides. So 30𝑅 sub 𝐶 is 5911.5. And when we divide both sides of this by 30, we get 𝑅 sub 𝐶 equals 197.05. But how does this help?

We were trying to find the value of the pressure 𝑃. But of course pressure is the amount of force acting on an area. So since this force is acting on a point, we can say that 𝑅 sub 𝐶 must actually be equal to 𝑃. And so 𝑃 must be equal to 197.05 newtons. Let’s clear some space and use this information in our first equation here. Using 𝑅 sub 𝐴 equals zero and 𝑅 sub 𝐶 as 197.05, our equation is as shown. We add 𝑤 to both sides and then subtract 166. And so we get that 𝑤 is 31.05 or 31.05 newtons.

We’re now going to recap the key points from this lesson. In this video, we saw that if a body is in equilibrium, we can say that the sum of all forces acting on that body must be equal to zero. Since force can have direction as well as magnitude, we define a direction for our forces and say that the sum of the forces in each direction is zero. We can also say that the sum of the moments must also be equal to zero. We saw that we can consider the forces and the moments acting on the body to help us create a system of linear equations, which we can solve to find missing values.

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