### 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.