### Video Transcript

In this lesson, we’ll learn how to
solve problems about the resolution of a force in two directions. We might recall that a force is an
interaction that can change the motion of an object. We often describe them as push or
pull forces, but there’s much more to them than that. We’re primarily interested in how
we can take a single force and resolve it into component — that means part —
forces. These components are often at right
angles to one another, but occasionally they might not be. And so we’re going to use right
angle trigonometry in addition to non-right angle trigonometry to help us to do
this. We’re going to begin with a very
simple example to see what this process might look like.

Resolve a force of 81 newtons into
two perpendicular components, 𝐹 sub one and 𝐹 sub two, as shown in the figure. Give your answer correct to two
decimal places.

And then we have a diagram that
shows a force of 81 newtons acting at an angle of 54 degrees to the horizontal. We’re also told that 𝐹 sub one and
𝐹 sub two, which are components of our 81-newton force, are perpendicular. This of course means that they meet
at an angle of 90 degrees, at a right angle. So how does this help us to
calculate the values of 𝐹 sub one and 𝐹 sub two?

Well, a nice method that we have is
to add in a right-angled triangle to our diagram. Since the two vertical lines are
parallel and 𝐹 sub two is itself a component of the 81-newton force, we can say
that the length of the vertical line that we added to our diagram must be equal to
the magnitude of 𝐹 sub two. And so since we have a right-angled
triangle for which we know the measure of one of its lengths and its angles, we can
use right angle trigonometry.

Our included angle is 54 degrees,
and our hypotenuse is 81 newtons. The side opposite to the angle
measuring 54 degrees is 𝐹 sub two, and the side adjacent to it is 𝐹 sub one. We’re going to begin by calculating
the measure of 𝐹 sub one. Since 𝐹 sub one is the length of
the adjacent side and we know the length of the hypotenuse, we’re going to use the
cosine ratio here. cos of 𝜃 is adjacent over
hypotenuse. And substituting everything we know
about our triangle into this formula, we get cos of 54 degrees equals 𝐹 sub one
over 81. We’ll solve for 𝐹 sub one by
multiplying through by 81. And so 𝐹 sub one is 81 times cos
of 54, which is 47.610 and so on. Correct two decimal places then, we
see that 𝐹 sub one is 47.61 newtons.

Let’s repeat this process for 𝐹
sub two. This time, we’re trying to find the
opposite side in our triangle. And so we’re going to use the sine
ratio, where sin 𝜃 is opposite over hypotenuse. Now, since at this point we already
knew two of the lengths, we could’ve used the Pythagorean theorem. But we’ll get the same answer
either way. This time, we get an equation sin
of 54 equals 𝐹 sub two over 81. Again, we solve for 𝐹 sub two by
multiplying both sides by 81. So 𝐹 sub two is 81 times sin of
54, which is 65.530 and so on. Correct two decimal places then, 𝐹
sub two is 65.53 newtons. And so we’ve resolved the force of
81 newtons acting at an angle of 54 degrees to the horizontal into two perpendicular
components.

Now, if we think about it, it does
make sense that the measures of 𝐹 one and 𝐹 two are smaller than 81. When we resolve a force, we’re
breaking it into its components, into parts. So 𝐹 sub one and 𝐹 sub two are a
part of the 81-newton force. And so they must be lower in
value.

In our next example, we’ll look at
how to resolve components of a force when they are not acting at right angles to one
another.

A force of magnitude 41 newtons
acts due south. It is resolved into two components
as shown on the diagram. Find the magnitudes of 𝐅 sub one
and 𝐅 sub two. Give your answer to two decimal
places.

Let’s begin by adding our force of
41 newtons to the diagram. It acts due south, so it’s this one
here. That force is itself then resolved
into two parts or components. Those are 𝐅 sub one and 𝐅 sub
two. Now, you might notice that these
two components are given in vector form. But actually, we’re trying to find
the magnitudes of these. The magnitude of the vector is
essentially its length. And so we want to find these two
dimensions here. And what we might try to do is add
right-angled triangles to our diagram.

But it’s not particularly easy to
do so here since the forces 𝐅 sub one and 𝐅 sub two are not acting at right angles
to one another. So instead, we draw something
called the parallelogram of forces. We’re drawing lines parallel to 𝐅
sub one and 𝐅 sub two, therefore creating a parallelogram. The force of 41 newtons acting due
south then makes up the diagonal of this parallelogram. And so we can then use angle facts
to complete some of our missing measurements.

We know that alternate angles are
equal, so we can add a 45-degree angle here and a 60-degree angle here. We also know that angles in a
triangle add up to 180 degrees. So by subtracting 45 and 60 from
180, we find the missing two angles in our triangles. Let’s enlarge the triangle on the
left-hand side, linking the force of 41 newtons and 𝐅 sub two. We have a non-right-angled triangle
for which we know the measure of all three angles and one of the sides. This means that to find the measure
of a second side, we can use the law of sines.

Labeling our triangle as shown, we
see we’re going to link 𝑎 over sin 𝐴 with 𝑏 over sin 𝐵. Substituting what we know about our
triangle into the formula, and we see that the magnitude of 𝐅 sub two divided by
sin of 60 is equal to 41 over sin of 75. To solve this equation to find the
magnitude of 𝐅 sub two — notice I’ve written that with bars — we multiply by sin of
60. And so the magnitude of 𝐅 sub two
is 41 over sin of 75 times sin of 60, and that’s equal to 36.759 and so on. Correct two decimal places then,
the magnitude of 𝐅 sub two is 36.76 newtons.

We’re going to repeat this process
to find the magnitude of 𝐅 sub one. Now, to do this, we could repeat
the process of redrawing the triangle on the right-hand side. However, we know that this side,
the side labeled 𝐴𝐵, on the triangle we drew is parallel to the side that
represents the vector component 𝐅 sub one. In fact, since this is a
parallelogram, it’s also of equal length. And so if we calculate the length
of line segment 𝐴𝐵, that will tell us the magnitude of 𝐅 sub one. This time, we’re linking 𝑏 and sin
𝐵 with 𝑐 and sin 𝐶. And so our equation is 41 over sin
75 equals the magnitude of 𝐅 sub one over sin of 45.

To solve for 𝐅 sub one, we’re
going to multiply both sides by sin of 45 degrees. And so the magnitude of 𝐅 sub one
is 41 over sin of 75 times sin of 45, which is 30.014 and so on. Correct to two decimal places,
that’s 30.01. And so the magnitude of 𝐅 sub one
is 30.01 newtons and the magnitude of 𝐅 sub two is 36.76 newtons.

Now, a key skill is to be able to
take the information in the question and construct your own force diagram. Let’s see what that looks like when
we’re working with another parallelogram of forces.

The angle between two forces, 𝑎
one and 𝑎 two, is 75 degrees. Their resultant is 2,900 newtons
and makes an angle of 45 degrees with 𝑎 one. Find the forces 𝑎 one and 𝑎
two. Give your answers to two decimal
places.

Before we do anything, we’re going
to begin by drawing a force diagram. Here are our two forces 𝑎 one and
𝑎 two, which meet at an angle of 75 degrees. Then, we have this resultant force
of 2,900 newtons that makes an angle of 45 degrees with 𝑎 one. Remember, the resultant force is
just the single force that we get by combining a system of forces, so here 𝑎 one
and 𝑎 two. Our job is to find the forces 𝑎
sub one and 𝑎 sub two. And so we’re going to add a couple
of things to our diagram.

Firstly, we add in lines parallel
to those representing the forces 𝑎 one and 𝑎 two. This gives us a parallelogram. And so we know that not only are
the opposite sides parallel, but they’re also equal in length. And that’s going to be useful later
down the line. Then we use the fact that alternate
angles are equal. And this angle must be 45
degrees. Then, by subtracting 45 from 75, we
get two angles of 30 degrees. The third angle is found by
subtracting 45 and 30 from 180 degrees, since there are 180 degrees in a
triangle. And so we see that the third angle
in both our triangles is 105 degrees.

Let’s separate out our
triangles. And we see that we’re dealing with
a pair of identical non-right-angled triangles. We have a length of 2,900 in both
triangles. And then we have 𝑎 sub one
here. And since we said that opposite
sides in a parallelogram are parallel and equal in length, we have 𝑎 sub two over
here.

Since we know the measure of all
the angles in our triangle and one of its lengths, we can use the law of sines to
find the other two. Let’s label our triangle as
shown. And to begin with, we’ll work out
the value of 𝑎 sub one. This means we want 𝑎 over sin 𝐴
equals 𝑐 over sin 𝐶. Substituting everything we know
about our triangle into this formula gives us 𝑎 sub one over sin of 30 degrees
equals 2,900 over sin of 105. We’re going to multiply both sides
by sin of 30 to find the value of 𝑎 sub one. And so 𝑎 sub one is 2,900 over sin
of 105 times sin of 30, which correct to two decimal places is 1,501.15. So 𝑎 sub one is 1,501.15
newtons.

Let’s clear some space and perform
the same process to find 𝑎 sub two. This time, we’re going to use 𝑏
over sin 𝐵 equals 𝑐 over sin 𝐶. Substituting into the formula, and
we see that we can solve for 𝑎 sub two by multiplying both sides by sin of 45. And 𝑎 sub two then is 2,900 over
sin of 105 times sin of 45. And that, correct to two decimal
places, is 2,122.95. 𝑎 sub one is 1,501.15 newtons, and
𝑎 sub two is 2,122.95 newtons.

In our final example, we’ll look at
how we can resolve the weight of the body that’s resting on an inclined plane into
two components in two directions.

A body weighing 72 newtons is
placed on a plane that is inclined at 45 degrees to the horizontal. Resolve its weight into two
components 𝐹 sub one and 𝐹 sub two, where 𝐹 sub one is the component in the
direction of the plane and 𝐹 sub two is the component normal to the plane.

Being able to model bodies resting
on a plane is going to be really important when it comes to working with Newton’s
laws of motion. So to see how it works, we’re going
to begin by drawing a sketch of the body resting on a plane. Here is a body resting on a plane
inclined at an angle of 45 degrees. The body has a weight of 72
newtons. And that’s essentially the
downwards force that it exerts on the plane due to gravity.

Now, usually, we’d look to add any
extra forces to the diagram. But actually, we’re just interested
in resolving this 72-newton force into two components, 𝐹 sub one and 𝐹 sub
two. We’re actually told that 𝐹 sub one
is the component in the direction of the plane. Essentially, it’s the component of
the weight that acts parallel to the plane. Then, 𝐹 sub two is the component
normal to the plane. It’s the component of our weight
that’s perpendicular to the plane. And so we can add these two forces
in as shown.

Now, in fact, you might be used to
seeing 𝐹 sub one here. This is parallel and of the same
length as the line that we’ve drawn on our triangle. And so 𝐹 sub one will be the same
value in either position. Now, in fact, 𝐹 sub one and 𝐹 sub
two are perpendicular to one another. And so we have a right-angled
triangle in which we know the length of the hypotenuse. That’s 72 newtons. This means then that if we can find
the measure of one other angle in our triangle, we can use right angle trigonometry
to resolve our weight into the components 𝐹 sub one and 𝐹 sub two. In other words, we can find
expressions and values for 𝐹 sub one and 𝐹 sub two.

Now, to find another angle then,
we’re going to draw a slightly larger right-angled triangle. In this triangle, we see that we
have a right angle at the point where the body meets the plane. We also have an angle of 45
degrees. This actually means we have an
isosceles right triangle. But we could also work out the
other angle by calculating 180 minus 90 minus 45. We see that this angle here is 45
degrees.

Now, we know that the 72 newtons
acts vertically downwards. And so this force must be acting at
an angle of 90 degrees to the horizontal. This means we can work out this
angle here. We have a right-angled triangle
with one angle of 90 and another of 45. And so 180 minus 90 minus 45 is 45
degrees. And so we now know one other angle
in the right-angled triangle we were interested in. In fact, it will always be the case
that the angle at which the plane is inclined to the horizontal will be the same as
this angle here. So let’s enlarge the triangle we’re
looking at to make it a little bit easier to work out what’s happening.

We now have a right-angled triangle
for which we know one of the sides and the measure of one of its angles. We’re going to use right angle
trigonometry to find the value of 𝐹 one and 𝐹 two. This 72-newton length is the
hypotenuse. Then, 𝐹 sub one is the opposite,
and 𝐹 sub two is the adjacent. To find 𝐹 sub one, we’ll use the
sine ratio. We know that 𝐹 sub one is the
opposite, and we have the hypotenuse. We can therefore say that sin of 45
degrees must be equal to 𝐹 sub one over 72. And to find the value of 𝐹 sub
one, we’ll multiply both sides by 72. That gives us that 𝐹 sub one is 72
times sin of 45.

But actually, we can quote that sin
of 45 is root two over two. So this becomes 72 root two over
two, which is 36 root two, and that’s in newtons. We then have two possible options
for calculating the value of 𝐹 sub two. We could use the cosine ratio. If we went down this route, we’d
find that 𝐹 sub two is 72 times cos of 45, which is also 36 root two or 36 root two
newtons. Alternatively, we might have
spotted that we were working with an isosceles right-angled triangle. And so 𝐹 sub one and 𝐹 sub two
absolutely had to be the same length. 𝐹 sub one and 𝐹 sub two are 36
root two newtons.

We might even choose to use the
Pythagorean theorem to double check our answer. By summing the squares of 36 root
two and 36 root two and then finding the square root of that value, we’d then get a
value of 72 newtons as expected. Now, we can generalize this idea a
little bit, although it is useful to be able to see where these values come
from. For a body resting on a plane
inclined at an angle of 𝜃 to the horizontal with a weight of 𝑤 newtons, the
component of the weight that acts parallel to the plane will always be 𝑤 sin 𝜃,
whereas the component of the weight that acts perpendicular to the plane will always
be 𝑤 cos 𝜃.

Let’s now recap the key points from
this lesson. In this video, we saw that to
resolve a force into component parts where those components are perpendicular to one
another, we can add right-angled triangles to our diagram and use right angle
trigonometry. This process is particularly useful
when working with bodies on inclined planes. When this isn’t possible, however,
that is, when the components are not at right angles to one another, we use a
diagram called a parallelogram of forces. Then, we use angle facts and the
sine rule to help us calculate the individual components.