# Lesson Video: Equilibrium of a Rigid Body Mathematics

In this video, we will learn how to solve problems about the equilibrium of rigid bodies in 2D where the sum of forces and the sum of moments equal zero.

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### Video Transcript

In this lesson, we’ll learn how to solve problems involving the equilibrium of rigid bodies in two dimensions, where the sum of forces and the sum of moments equal zero. We’re going to see that for a rigid body or object, such as the various parts of this suspension bridge, when the total force and the total moment of force acting on the body is zero, that body is said to be in equilibrium. And these are conditions we’ll be able to use to analyze different examples. So, what do we mean when we talk about a rigid body?

A rigid body is an object that doesn’t bend or flex or change shape in any way due to the action of a force or a number of forces. A rigid body is said to be in equilibrium if the sum of the forces and the sum of the moments on the body are zero. This means that the body does not experience either transitional motion or rotational motion. Consider, for instance, a uniform rod at rest on a smooth horizontal plane. The rod exerts a weight force vertically downward on the table. By Newton’s third law of motion, the table exerts a normal reaction force 𝑅 on the rod. Since these forces are equal in magnitude and opposite in direction and there are no other forces acting on the body, 𝑅 is equal to 𝑊, meaning that the net sum of the forces is zero. In turn, we’re able to deduce that the acceleration of the body is zero and it is in transitional equilibrium.

Now, what about rotational equilibrium? Well, if the forces act along the same line of action through our object, this means if we think about the tendency of either one of these forces to make this rod rotate to create a nonzero moment, then we can see regardless of the point we choose as a potential axis of rotation so long as that point is along the rod’s axis, then the moment of the force created by either one of our forces about that point will be exactly counterbalanced by the moment created by the other force. When the net moment on the body is zero, we say it’s in rotational equilibrium. And when these two conditions are both met, we say that it’s simply in equilibrium.

Let’s formalize this. A body under the action of a set of coplanar forces is in static equilibrium if the resultant force vanishes — that is, the sum of the vector forces is equal to zero — and the resultant moment about any arbitrary point, let’s say 𝑝, also vanishes — that is, the sum of the vector moments is equal to zero. These conditions are sufficient and necessary for the equilibrium of the forces. Now let’s write the resultant force as 𝑥𝐢 plus 𝑦𝐣, where 𝐢 and 𝐣 are coplanar perpendicular unit vectors.

It follows that if 𝑥 is equal to zero and 𝑦 is equal to zero, the sum of the forces is equal to zero. In addition to this, if the sum of the algebraic measures of the moments of the forces about a point in the plane vanish, then the body is said to be in equilibrium. And this is really useful because it means we can extend the concepts of working with the body in equilibrium using vector notation to allow us to work with scalars. Let’s demonstrate what this means in practical terms for solving problems involving rigid bodies in equilibrium.

In the given figure, determine the magnitude of the force 𝐅 that makes the rod in equilibrium, given that the magnitude of the given force is seven newtons and cos 𝜃 is equal to four-fifths.

So, we have a vertical rod, and we’re told that this is in equilibrium. It has two known forces acting on it, but these are the only forces. Consider the point where the rod meets the surface that it rests on; that’s point 𝐴. There’s a downward-acting weight force exerted on this point. But by Newton’s third law of motion, there’s also an oppositely directed reaction force; let’s call that 𝑅. And along with these forces, there is indeed one more force that acts on the rod. We know this is true because consider the rotation of the rod about either point where the two horizontal forces act.

Considering only the forces we drawn in so far, we can see the rod won’t be in rotational equilibrium; that is, it will have a nonzero moment. This issue is resolved if we assume some frictional force, let’s call that 𝐅 sub 𝑟, acting on our rod to the left at point 𝐴. And now, we have a free-body diagram showing all the forces acting on our rod.

Now we know that for the body to be in equilibrium, the sum of the forces must be equal to zero and the sum of the moments about some point — let’s call that 𝑃 — must also be equal to zero. Now, we’ve written this using vector notation. But of course, we know that we can simply resolve horizontally and vertically and achieve the same result. Because there are several unknown forces here, we’re going to focus on the sum of the moments. And we’re going to form an equation involving them, taking the sum of the moments about point 𝐴.

Since the moment is force times perpendicular distance from the point 𝐴 to the line of action of the force, the moment of our seven-newton force is seven cos 30 times 4.7 plus 2.1. That’s 47.6 cos 30. Similarly, the moment of the unknown force 𝐅 will be 𝐅 cos 𝜃 times 2.1. Next, we’ll take the sum of the moments, taking the counterclockwise direction to be positive. And since we know that sum is equal to zero, we obtain 47.6 cos 30 minus 2.1𝐅 cos 𝜃 equals zero.

In fact, we haven’t yet used the information that cos 𝜃 is equal to four-fifths. So, let’s substitute that into our equation, and we’ll be able to solve for 𝐅. When we do, the second part of our expression becomes negative 2.1𝐅 times four-fifths, which is 1.68𝐅. Rearranging for 𝐅 from we have, it’s equal to 47.6 cos 30 divided by 1.68. That gives us 85 over three times root three over two, which simplifies to 85 root three over six newtons. And so, we’ve obtained the magnitude of the force 𝐅 that makes our rod in equilibrium. It’s 85 root three over six newtons.

In this example, we could solve missing force problems by just considering the moments. In our next example, we’re going to look at a ladder problem. It’s an example where a ladder is resting against a smooth wall with its base on rough horizontal ground.

A uniform ladder is resting in a vertical plane with its upper end against a smooth vertical wall and its lower one on a rough horizontal floor, where the coefficient of friction between the ladder and the floor is two-thirds. The ladder is inclined to the horizontal at an angle measuring 48 degrees. Given that the ladder weighs 295 newtons and has a length 𝐿, find in terms of 𝐿 the maximum distance a man weighing 610 newtons can climb up the ladder without it slipping, rounding your answer to two decimal places.

There’s an awful lot going on here, so we’re going to begin by drawing a free-body diagram. Here is the uniform ladder resting against a smooth vertical wall at an angle of 48 degrees to the horizontal. Since the ladder is uniform, we assume that its weight acts vertically downwards at a point exactly halfway along the ladder. Then, there are reaction forces of the floor on the ladder and the wall on the ladder; we’ll call them 𝑅 sub 𝐴 and 𝑅 sub 𝐵, respectively. The ladder is stopped from slipping by the frictional force. Let’s call that 𝐅 sub 𝑟. But of course, that’s equal to 𝜇 times 𝑅 sub 𝐴, where 𝜇 is the coefficient of friction and 𝑅 sub 𝐴 is the reaction force of the floor on the ladder.

Finally, we know that the man can walk some distance up the ladder without it slipping. He weighs 610 newtons, so we’ll model that as being a little bit past the halfway point. In fact, if we model the weight of the ladder to be a half 𝐿 from the base of the ladder, then we can say that the man is 𝑥𝐿 from the base of the ladder, where 𝑥 is some fraction. And we can see we’re actually trying to work out the value of 𝑥. So, we can clear some space and start solving this problem.

Now at the exact instant before the ladder slips, it will be on the point of slipping, so it’s going to be in equilibrium. Now for this to be true, we know that the net force acting on the body must be zero, but also the net moments must also be zero. In fact, rather than working in vector form, we’ll resolve individual components for the force. Let’s choose downwards to be positive and find the net sum of all forces that act on the ladder.

There are three forces that act in this vertical direction. They are the weight of the ladder, the weight of the man, and the reaction force 𝐴 of the floor on the ladder. Their sum then is 610 plus 295 minus 𝑅 sub 𝐴. Since the ladder is in equilibrium, that’s equal to zero. We can therefore say that 905 minus 𝑅 sub 𝐴 is equal to zero. And 𝑅 sub 𝐴, the reaction force of the floor on the ladder, is therefore equal to 905 newtons. Let’s now resolve horizontally, and we’ll take the right to be positive.

The sum of the forces in this direction is 𝑅 sub 𝐵 minus the frictional force. And we subtract that frictional force because it’s acting to the left. And of course, the sum of these forces is equal to zero. This means, in fact, that the reaction force at 𝐵 must be exactly equal to the friction force. But remember, we said the frictional force is equal to the coefficient of friction times the normal reaction force at that point, so 𝜇 times 𝑅 sub 𝐴.

We calculated the reaction force at 𝐴 to be equal to 905. And we were told in the question that 𝜇, the coefficient of friction, is two-thirds. Two-thirds times 905 is 1810 over three. And so, we have the reaction force 𝐵. That is the reaction force of the wall on the ladder. With this in mind, we’re now able to take moments about some point. Now, we can take moments about any point, but it often makes sense to take them about the foot of the ladder, since there’s more than one force acting here, and it can help us to simplify things. Let’s take moments about point 𝐴 then.

We know that we’re aiming for the sum of the moments about point 𝐴 to be equal to zero, and we’re also going to say that the counterclockwise direction is positive. Now, at this stage, it’s worth noting that the 610 force and the 295-newton force are not acting in a direction perpendicular to the ladder. So, we’re going to resolve these forces. Using right-triangle trigonometry, we see that the component of these forces that is perpendicular to the ladder is 610 cos 48 and 295 cos 48.

The moment of the weight of the man about this point is the product of this component of the force and the distance from point 𝐴. So, that’s 610 cos 48 times 𝑥𝐿. Similarly, the moment of the weight of the ladder is 295 cos 48 times half 𝐿. In the opposite direction, we have the moment of the reaction force of the wall against the ladder. So, we subtract 1810 over three sin 48 times 𝐿. Since the sum of these forces is equal to zero, we set the sum equal to zero. We now observe that we can divide through by 𝐿. And we do that because we know that the length of the ladder cannot be equal to zero.

Next, we look to make 𝑥 the subject. So, we subtract 295 over two cos 48 from both sides and add 1810 over three sin 48. Finally, we divide through by 610 cos of 48 degrees, and calculating this will give us the value of 𝑥. This gives us 0.85667 and so on. And so, in terms of 𝐿, the distance that the man can climb the ladder before it slips is 0.86𝐿 or 0.86𝐿 units.

Let’s now recap some of the key points from this lesson. In this video, we’ve learned that a rigid body is an object that doesn’t bend or flex or change shape in any way due to the action of a force. A body under the action of a set of coplanar forces is in static equilibrium if the resultant force vanishes — that is, the sum of the forces is zero — and the resultant moment about any arbitrary point, say 𝑃, also vanishes. These conditions are sufficient and necessary for the equilibrium of forces. And remember, we saw that we can extend the concepts of a body in equilibrium using vector notation to allow us to work with scalars, and this can be more efficient for our calculations.