Video: Equilibrium of a Rigid Body | Nagwa Video: Equilibrium of a Rigid Body | Nagwa

# Video: Equilibrium of a Rigid Body

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.

18:06

### Video Transcript

In this video, we’re talking about equilibrium of a rigid body. 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 a given body is zero, that body is said to be in equilibrium. These are conditions we’ll be able to use to analyze different examples.

As we get started, we can define a rigid body as an object that doesn’t bend or flex or change shape in any way. And equilibrium means that such a body is not experiencing either translational motion or rotational motion. As an example, we can consider this rod at rest on the ground. The only forces acting on this rod, its weight force acting down and a reaction or normal force that’s acting upward, effectively cancel one another out so that the net force experienced by our rod is zero. This tells us that this object is not accelerating and therefore is in translational equilibrium. As we mentioned, though, that’s just one aspect of being in equilibrium.

Considering once more the forces acting on this rod resting on the ground, we can see that they 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 — in other words, to create a nonzero moment about some point — then we can see that 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 force, or moment for short, created by either one of our two forces about that particular point will be exactly counterbalanced by the moment created by the other force.

For example, say that we choose as a rotation axis the point on the left most end of this rod. Knowing that in general a moment of force is equal to the perpendicular component of a given force multiplied by the distance between its point of action and a possible axis of rotation, we can see that for our rod that distance for both of our forces is this distance here. That is, it’s the same in each case. And since the reaction force 𝑅 is equal and opposite to the weight force, we can say that whatever moment the force 𝑅 tends to create around this point is counteracted by the moment created by force 𝑤.

This, as it turns out, is the second condition for equilibrium. When the net moment on a rigid body is zero, that means it’s in rotational equilibrium. And when these two conditions are both met, we say that it’s simply in equilibrium. That’s true for our rod. And note that this is the case regardless of the point along the rod’s axis that we choose to consider rotations about. These are the two conditions we’ll rely on to answer questions about rigid bodies and equilibrium. Once we’re told that a given object is in equilibrium, we know that these are both true.

Before we look at a couple of example exercises, let’s outline a solution approach we can take for rigid bodies in equilibrium. Given a scenario like this, our first step can be to draw a free body diagram, one that shows all the forces acting on our object. In the case of our rod, we already have all those forces sketched in. Our next step is to set up sign conventions, both for forces and moments. We might say, for example, that a force directed vertically upward is in the positive direction, and one directed downward is therefore negative. And since vertical and horizontal forces are independent of one another, it may also be helpful to define positive and negative in that second dimension. Along with positive and negative forces, we can have positive and negative moments.

For moments, let’s set up the convention that a moment that tends to make an object rotate counterclockwise is positive. So going back to our axis of rotation about the left most point on our rod, this would mean the moment created by the reaction force 𝑅 would be positive, while that created by the weight force would be negative. A third and related step in our solution approach is to establish a coordinate frame. We may say, for example, that forces to the right are in the positive 𝑥-direction, and those directed upward are in the positive 𝑦.

Our last step, and it’s a big one, is to establish and then solve systems of equations for unknown values. We get these systems of equations from these conditions for equilibrium. For example, condition one might lead to two independent equations, one for vertical forces and one for horizontal. Condition two may yield a third independent equation, which means that once we write up our system, we may be able to solve for as many as three unknowns. Well, the best way to see how all this works is through practice. So let’s look now at an example exercise.

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 of 𝜃 equals four-fifths.

Looking at our figure, we see this vertical rod, which we’re told is in equilibrium. Two forces are drawn in acting on the rod, one here of magnitude seven newtons and an unknown force 𝐹. We know, though, that these aren’t the only forces that act on this rod. If we consider the point where the rod meets the surface that it rests on, this is labeled as point 𝐴. We know that there’s a downward acting weight force being exerted on this point, and there’s also an oppositely directed reaction force; we’ll call it 𝑅. Note that the magnitude of this reaction force is greater than that of 𝑤 because it needs to counteract the downward acting components of the seven-newton force and the unknown force 𝐹.

Along with these forces, there’s actually one more that acts on this rod. We know this because if we were to consider the rotation of the rod about either this point here or this point here, where one of our two horizontal forces act, then considering only the forces we’ve drawn in so far, we can see that actually this rod would not be in rotational equilibrium. That is, it would have a nonzero moment. This issue is resolved if we assume a frictional force, we’ll call it 𝑧, acting on our rod to the left at point 𝐴. And now our free body diagram showing all the forces acting on our rod is complete.

Knowing this, we can recall the two conditions that must both be satisfied for a rigid body to be in equilibrium. First, the net force acting on the body must be zero. This ensures translational equilibrium. And then the net moment acting on a body in equilibrium must also be zero. This guarantees rotational equilibrium. Applying these conditions to our given situation will yield force and moment equations. For example, we can see that we have force components that are both vertical and horizontal, which would therefore yield two force balanced equations.

Furthermore, seeing that several of our forces would create a moment about the vertical axis of this rod, we can apply the second condition to yield another equilibrium equation. Let’s recall that it’s the magnitude of this unknown force 𝐹 that we want to solve for. We could start by applying either one of these equilibrium conditions. But notice that if we apply the second condition that the net moment of this rod must be zero and if we choose point 𝐴 as the place about which we calculate moments, then that choice of location will eliminate from consideration both the weight, the friction, and the reaction forces. And that’s because all three of these forces act at this point and therefore have a distance of zero between their line of action and this rotation axis.

That means if we’re considering the net moment of the forces on this rod about point 𝐴, we only need to take into account our seven-newton force, which acts on the rod at this location, and our unknown force 𝐹, which acts here. Recalling that the moment due to a certain force is equal to the distance from where that force is applied to the axis of rotation multiplied by the component of the force perpendicular to that line, we can see that we’ll be interested in the horizontal component of our seven-newton force and that of our unknown force 𝐹 along with the perpendicular distances from these forces to point 𝐴.

As we start writing out our moment balance equation that the net moment on this rod is zero, let’s set up the convention that any moment that leads to a counterclockwise rotation about point 𝐴 we’ll consider positive. This means that a moment causing rotation in the other direction is considered negative. Set up this way, we can see that the moment caused by the seven-newton force will be positive, while that caused by our unknown force 𝐹 will be negative. We can write that seven newtons times the cos of 30 degrees — by taking this cos, we isolate the horizontal component of the seven-newton force — multiplied by the distance from where this force is applied to point 𝐴 — which we can see is 4.7 meters plus 2.1 meters, that’s a total of 6.8 meters — minus, because it generates a moment we’ve called negative, the force magnitude 𝐹 times the cos of the unknown angle 𝜃 multiplied by 2.1 meters, that’s the distance from where 𝐹 is applied to the point 𝐴, is all equal to zero.

Just to be clear about why we’ve multiplied our force 𝐹 by the cos of 𝜃, notice that 𝜃 is this given angle here and that as an alternate interior angle, this is 𝜃 as well. Therefore, the horizontal component of 𝐹 is indeed 𝐹 times that cos of 𝜃. Now, in our problem statement, we’re told that the cos of 𝜃 is equal to four-fifths. And we can also recognize that the cos of 30 degrees is the square root of three over two. Making these substitutions, we’re now ready to rearrange and solve for the force 𝐹. If we add 𝐹 times four-fifths times 2.1 meters to both sides of this equation, then we get this result. Shifting this up a bit, let’s now divide both sides of this equation by four divided by five multiplied by 2.1 meters.

From here, we see that the units of meters cancel from numerator and denominator. And if we then multiply both numerator and denominator of our left-hand side by five divided by four, we then have this expression for the force magnitude 𝐹. This is equal to 35 divided by eight multiplied by 68 divided by 21 times the square root of three newtons. And if we multiply together and simplify as far as possible these two fractions, then we find they simplify to 85 divided by six, which means that the magnitude of the force 𝐹 is eighty-five sixths times the square root of three newtons.

Let’s now look at an example of an object in equilibrium inclined at an angle.

A uniform rod 𝐴𝐵 of weight 10 newtons and length 12.5 meters is resting with its end 𝐴 on a rough horizontal plane and point 𝐶, between 𝐴 and 𝐵, resting against a smooth horizontal nail, which is 5.7 meters above the horizontal plane. If the rod is about to slide when it is inclined to the horizontal at an angle whose tangent is three-quarters determine the coefficient of friction between the rod and the horizontal plane.

Looking at our figure, we see this rod with ends 𝐴 and 𝐵, where end 𝐴 is in contact with a rough horizontal plane and partway between the ends, the rod is supported by a smooth horizontal nail at point 𝐶. We want to solve for the coefficient of friction between the rod and the horizontal plane, which tells us that there’s a frictional force, we’ll call it 𝐹, acting in this direction on the rod at point 𝐴. We’re told further that the tangent of the angle between the horizontal plane and the rod is three-quarters, that the weight of the rod, we’ll call it 𝑤, is 10 newtons, and that its length overall, we’ll call that 𝑙, is 12.5 meters.

As we get started solving for the coefficient of friction involved in this frictional force, let’s clear some space on screen and remember the two conditions necessary for a rigid body like our rod to be in equilibrium. We’re told after all that this rod is just about to start sliding, meaning that currently it is in equilibrium. The first condition of equilibrium is that the net force on an object be zero. And the second is that the net moment acting on the object is zero. Both of these conditions rely on the forces acting on our object of interest. Let’s start on our solution, then, by drawing a free body diagram depicting all the forces acting on our rod.

We’ve already drawn in the frictional force that acts at point 𝐴. Also, at this point, there’s a vertically upward-acting normal or reaction force we’ll call 𝑅. And then we know at the rod’s midpoint, which we’ll say is here, there’s a downward acting force; that’s the weight force. And lastly, from the nail at point 𝐶, there’s a force acting on the rod, which is at right angles to the rod’s length. We know this because we’re told that the nail at point 𝐶 is smooth. Therefore, it can only exert a force on the rod in this particular direction. With our forces drawn in, let’s set up the sign conventions that forces that point to the right are in the positive 𝑥-direction and those that point upward are in the positive 𝑦.

Likewise, we can say that a moment created by any one of these forces will be considered positive if it tends to make the rod rotate counterclockwise. With these conventions in place, we can start to consider our first equilibrium condition. And because vertical and horizontal forces are independent of one another, this condition can actually generate two independent equations, one for the horizontally acting forces in our scenario and one for the vertically acting forces.

First, the horizontal forces. As we consider these, we see that the frictional force 𝐹 is directed in the positive horizontal direction. The only other horizontally acting force is a component of our force generated by the nail at point 𝐶. We’ll call this force magnitude 𝐹 sub 𝐶. And to solve for its horizontal component, the one indicated by our blue arrow, we’d like to know what angle this is right here in this right triangle showing the components of 𝐹 sub 𝐶. To figure that out, let’s recall that this angle here is 𝜃. Since this is one of the angles in a right triangle, we know that the remaining interior angle here must be 90 degrees minus 𝜃 since the sum of all three interior angles is 180 degrees.

But then, this angle we’ve just indicated is identical to this angle here. And that, we can see, plus this interior angle that we actually want to solve for must be equal to 90 degrees. That’s because of the force 𝐹 sub 𝐶, as we saw, is perpendicular to the axis of the rod. So if this angle here is 90 degrees minus 𝜃, then that must mean that this interior angle is itself 𝜃. And therefore, the horizontal component of 𝐹 sub 𝐶 is 𝐹 sub 𝐶 times the sin of 𝜃. This allows us to complete our horizontal force balance equation.

Now, we’ll consider our vertical forces. From our free body diagram, we see that the reaction force 𝑅 is a positive vertical force, the weight force 𝑤 is a negative one, and that the force due to our nail 𝐹 sub 𝐶 also has a vertical component. In this case, it’s positive. And that component is 𝐹 sub 𝐶 times the cos of 𝜃. And all these forces added together, we know, must be zero because our rod is in equilibrium. Now that we have two independent equations, let’s consider again what we want to solve for in this exercise. Ultimately, we want to calculate 𝜇, the coefficient of friction. This coefficient of friction is equal to the maximum frictional force our horizontal surface can apply to the rod before the rod slips divided by the reaction force 𝑅.

So to solve for 𝜇, we’re going to need to know the reaction force we find here as well as the frictional force that we’ve called 𝐹. We see though that those aren’t the only unknowns in these equations. We also don’t know the magnitude of the force 𝐹 sub 𝐶. To figure all this out then, we’ll need to solve for three unknowns, meaning we’ll need another equation in our system of equations. And we can get that equation from applying the second condition of equilibrium. The net moment due to all these forces on the rod about any point along the rod’s axis must be zero.

And the nice thing is for any net moment calculation, we can choose any point along the rod’s axis for our rotation point. Let’s choose point 𝐴, where the rod meets the rough horizontal plane. Because the moment due to a given force is equal to the perpendicular component of that force multiplied by the distance from where the force acts to the rotation point, this choice of 𝐴 means that we don’t have to consider the reaction force or the frictional force in our calculation of moment. That’s because for each of these forces, the distance to 𝐴 is zero, knowing out that moment.

Remembering we established that any moment tending to create a counterclockwise rotation about our point of interest is positive, we can see that the moment created by 𝐹 sub 𝐶 will be positive, while that created by the weight force will be negative. The moment due to 𝐹 sub 𝐶 is equal to 𝐹 sub 𝐶 times this distance here. Now, since the tangent of 𝜃 is equal to three-quarters, if we consider the right triangle, now with sides in orange, we see it’s an example of a three, four, five triangle, which means that the hypotenuse of this triangle, the distance we’re interested in, is 5.7 meters times five-thirds.

From this moment, we subtract the component of the weight force perpendicular to the rod’s axis, noting that in this right triangle of its components, this angle here is 𝜃, which means that component is 𝑤 times the cos of 𝜃. And we multiply this by the length of the rod over two. And because of equilibrium, this sums to zero. Knowing 𝑤, that the cos of 𝜃 is four-fifths, and 𝑙 is 12.5 meters, we can now solve this moment balance equation for the force 𝐹 sub 𝐶. When we do, we find it’s one hundred nineteenths, which we can then substitute into our horizontal force equation to solve for 𝐹 and our vertical force equation to solve for 𝑅. 𝜇 is equal to 𝐹 max over 𝑅. And because the rod is on the point of slipping, we know that the friction force 𝐹 is equal to 𝐹 max. So we can use this value to find that 𝜇 is equal to six elevenths. . This is the coefficient of friction.

Let’s summarize now using a few key points. In this lesson, we learned to conditions that a rigid body in equilibrium must satisfy. And lastly, you learned a solution process for equilibrium exercises.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions