### Video Transcript

A ladder 𝐴𝐵 weighing 40 root
three kilogram-weight and having a length of five meters is resting in a vertical
plane with end 𝐵 on a smooth floor and end 𝐴 against a smooth vertical wall. End 𝐵 is attached by a string to a
point on the floor vertically below 𝐴. Given that 𝐵 is 2.5 meters away
from the wall and the weight of the ladder is acting at a point on the ladder two
meters away from 𝐵, find the tension in the string.

Before we perform any calculations,
we’re going to begin by drawing a free body diagram. This is a very simple sketch of the
scenario that shows all of the forces we’re interested in. Here is a ladder with a length of
five meters resting against the smooth vertical wall. We don’t know the angle that the
ladder makes with the horizontal, but we are going to need to calculate it at some
point. So let’s call that 𝜃. We’re told that end 𝐵 is attached
by a string to a point on the floor vertically below 𝐴. This means we need to add a
tensional force 𝑇 to the diagram as shown. Then we’re told that 𝐵 is 2.5
meters away from the wall. Now this will be useful in a moment
cause we’re going to be able to use right angle trigonometry to calculate the value
of 𝜃.

We’re then told that the weight of
the ladder acts at a point on the ladder two meters away from 𝐵. So that’s a downwards force of 40
root three kilogram-weight. There are two further forces that
we’re interested in. We know by Newton’s third law of
motion that since the ladder will be exerting a force on the floor and the wall,
there will be normal reaction forces of the floor and the wall on the ladder,
respectively. Let’s call them 𝑅 sub 𝐵 and 𝑅
sub 𝐴. We now have all the key information
and we’re trying to find the tension in the string. We’ve called that 𝑇. So let’s clear some space.

So what do we do next? Well, we know that the ladder is
resting in its position. And so we can assume that it’s in
equilibrium. For a rigid body such as this
ladder to be in equilibrium, there are two conditions that need to be met. Firstly, the sum of all the forces
acting on the body must be equal to zero. We often break these down into
horizontal and vertical forces such that the sum of the horizontal forces is zero
and the sum of the vertical forces is zero. We’re also able to say that the sum
of all the moments acting on the body must be equal to zero, where moment is
calculated by finding the product of the force 𝐹 and the distance 𝑑, where 𝑑 is
the perpendicular distance of the line of action of the force from the point about
which the object is trying to rotate.

As with forces, we also give a
direction to our moments. This time those are clockwise and
counterclockwise. So let’s begin by finding the sum
of the forces on our diagram. We’re trying to calculate the
tension. And we really aren’t interested in
𝑅 sub 𝐵 at all. So let’s begin by calculating the
sum of the horizontal forces. Let’s take the direction in which
the tension force is acting to be positive. Then we have 𝑅 sub 𝐴 acting in
the other direction. So the sum of the forces in a
horizontal direction is 𝑇 minus 𝑅 sub 𝐴. And that, of course, is going to be
equal to zero.

Next, we’re going to take moments
about 𝐵. Now we can take moments about any
point on the ladder, although we often choose the foot of the ladder since there are
more forces acting there. Now, in fact, there are two forces
that we’re going to be looking at, and that is the downwards force of the weight and
the reaction force at 𝐴. But remember, we said the moment
was calculated by multiplying the force with its perpendicular distance from the
pivot. And so we’re going to need to
resolve these forces into their components which act perpendicular to the
ladder. We can add right-angled triangles
as shown with an included angle of 𝜃. So we actually need to calculate
the value of 𝜃 next.

By enlarging just the outline of
the floor, the wall, and the ladder, we see we can use right angle trigonometry. We have an adjacent side of 2.5
meters and a hypotenuse of five. And so we use the cosine ratio. cos of 𝜃 is adjacent over
hypotenuse. In our triangle, that’s cos of 𝜃
is 2.5 divided by five. But 2.5 divided by five is equal to
one-half. And since we know cos of 60 is
equal to one-half 𝜃 must be equal to 60 degrees. We can therefore add 60 degrees in
place of 𝜃. And we’re ready to start
calculating the components of our weight and reaction force that are perpendicular
to the ladder.

Let’s begin with the component of
the weight. We enlarge this triangle a little
bit. Let’s call the component we’re
trying to find 𝑥. 𝑥 is the adjacent side in our
triangle, whereas the hypotenuse is this 40 root three kilogram-weight. We can therefore use the cosine
ratio once again to find the value of 𝑥. We get cos of 60 is 𝑥 over 40 root
three. But remember, cos of 60 degrees was
one-half. So let’s solve this equation by
multiplying both sides by 40 root 3. One-half times 40 root three is 20
root three. So the component of the weight
force that acts perpendicular to the ladder is 20 root three kilogram-weight.

So what is the moment of this
force? We’re going to say it’s negative,
since we chose counterclockwise to be positive. It’s negative 20 root three times
the distance from that pivot; that’s two. We’re now going to repeat this
process for 𝑅 sub 𝐴. Let’s call the component of this
force that we’re trying to find 𝑦. This time, 𝑦 is the opposite side
in our triangle. And we have an expression for the
hypotenuse. The trigonometric ratio that links
these is the sine ratio. So let’s substitute what we know
about our triangle into this formula.

When we do, we get sin of 60 is
equal to 𝑦 over 𝑅 sub 𝐴. In fact, though, sin of 60 is root
three over two. So let’s multiply both sides of
this equation by 𝑅 sub 𝐴 to find an expression for 𝑦. When we do, we get 𝑦 is equal to
root three over two times 𝑅 sub 𝐴. Now, before we calculate the
moment, let’s go back to that first equation we wrote. If we add 𝑅 sub 𝐴 to both sides,
we find that 𝑇 is equal to 𝑅 sub 𝐴. Now this is really useful because
if we replace 𝑅 sub 𝐴 with 𝑇, we find 𝑦 is equal to root three over two 𝑇. And when we complete the equation
that shows the sum of our moments is equal to zero, we’re going to have an equation
purely in terms of 𝑇.

The moment of this force, then, is
root three over two 𝑇 times the distance from the pivot, so that’s five. And of course, we know that sum of
these moments is zero. We need to solve this equation for
𝑇, so the first thing we’re going to do is just divide through by root three. When we do, our equation becomes
negative 40 plus five over two 𝑇 equals zero. We can add 40 to both sides, and
our final step will be to divide through by five over two. This is the same as dividing by
five and multiplying by two. 40 divided by five over two then is
16. And so we’ve calculated the tension
in the string. We know our units for force in this
question are kilogram-weight. So the tension is 16
kilogram-weight.