Video Transcript
In this video, we’ll learn how to
find the resultant of a group of forces acting on a point. Now, we might recall that there are
many ways in which forces can be manipulated. And it’s often easier to work with
a large complicated system of forces by reducing it to a number of smaller
problems. We call this process the resolution
of forces or force systems. And it’s one way to simplify what
may otherwise seem to be an impossible system of forces acting on a body.
In this video, we’re primarily
interested in coplanar force systems. Now, when we talk about this type
of force system, we’re talking about a system in which the forces act in just one
plane. Now, those forces might be parallel
or perpendicular to one another, or even acting at an angle. Now, in those cases, we’ll need to
use trigonometry to resolve them into their various components, usually in terms of
the horizontal and vertical components. Then, once we’ve done that and
defined a positive direction of course, we’re able to find the resultant of our
forces. And that’s just the sum of the
separate forces.
So we’re going to begin by just
looking at an example of how this works. And we’re going to consider how to
apply right angle trigonometry to a system of forces given in vector notation.
The resultant of forces 𝐅 sub one
equals negative four 𝐢 plus two 𝐣 newtons, 𝐅 sub two equals five 𝐢 minus seven
𝐣 newtons, and 𝐅 sub three equals two 𝐢 plus nine 𝐣 newtons makes an angle 𝜃
with the positive 𝑥-axis. Determine 𝑅, the magnitude of the
resultant, and the value of tan of 𝜃.
We’re given three vector forces in
this question. These vector forces are given in
terms of 𝐢- and 𝐣-components. 𝐢 and 𝐣 are unit vectors, and
they act perpendicular to one another. And we generally say that the unit
vector 𝐢 acts in the positive 𝑥-direction, whilst the unit vector 𝐣 acts in a
positive 𝑦-direction.
We’re actually being asked to
calculate some information about the resultant of these forces. And we recall of course that the
resultant of a number of forces is simply the vector sum of those forces. So we’re going to add together the
vector 𝐅 sub one, 𝐅 sub two, and 𝐅 sub three. We can therefore say that the
resultant of our three forces is going to be a single vector in terms of 𝐢 and
𝐣. We calculate this vector by adding
negative four 𝐢 plus two 𝐣, five 𝐢 minus seven 𝐣, and two 𝐢 plus nine 𝐣.
Now, of course, to add a number of
vectors, we simply add their individual components. So we’ll begin by looking at the
𝐢-components. That’s negative four 𝐢 plus five
𝐢 plus two 𝐢. Negative four plus five plus two is
three. So we know the 𝐢-component of our
resultant is three. Then we add the 𝐣-components,
that’s two, negative seven, and nine, giving us a 𝐣-component of four for our
resultant vector. And of course this is in newtons
still. And so that’s the resultant of our
forces.
But in fact, we were looking to
find 𝑅, the magnitude of the resultant. And we might recall that, as a
direct result of the Pythagorean theorem, the magnitude of a vector whose
𝐢-component is 𝑎 and whose 𝐣-component is 𝑏 is the square root of the sum of the
squares of the individual components. So it’s the square root of 𝑎
squared plus 𝑏 squared. And this, therefore, means that the
magnitude of our resultant will be the square root of three squared plus four
squared. Well, three squared plus four
squared is 25. So we’re finding the positive
square root of 25, which is simply five. The resultant vector is in newtons,
and so the magnitude of this vector must also be in newtons. And we can, therefore, say that the
magnitude of the resultant is five newtons.
We still need to find the value of
tan of 𝜃, where 𝜃 is the angle that our resultant makes with the positive
𝑥-axis. Let’s sketch this out and just see
what it might look like. Here is our resultant vector. It’s three 𝐢 plus four 𝐣. Since the horizontal component, the
𝐢-component, is three units and the vertical component 𝐣 is four units, we can add
a right-angled triangle with the measurements three and four units as shown. Of course, we’ve worked out the
resultant, and that’s equal to five newtons. 𝜃 is this angle here. It’s the angle that our resultant,
the yellow line, makes with the positive 𝑥-axis.
We have a right-angled triangle for
which we actually know three of its sides. And so to find the value of the
included angle, we’re going to use right angle trigonometry. We then see that the side adjacent
to the included angle is three units and the side opposite to it is four units. We can therefore use the tan ratio
to work out the value of 𝜃. Since tan of 𝜃 is opposite over
adjacent, here tan of 𝜃 must be equal to four-thirds. And in fact, we’re finished.
We could, if we were required to,
work out the value of 𝜃 by finding the inverse or arc tan of both sides of our
equation. But we were actually asked to find
the value of tan of 𝜃, and so we’re done. The value of 𝑅 is five newtons,
and tan of 𝜃 is equal to four-thirds.
So that’s great. We’ve seen how to find the
resultant of coplanar forces given in vector form. And we’ve also seen how to
calculate bits of information from that. But what do we do if we’re given a
diagram and forces acting at an angle? In our next example, we’ll find
out.
A body has a force of 10 newtons
acting on it horizontally, 25 newtons acting on it vertically upward, and five
newtons acting on it at an angle of 45 degrees to the horizontal, as shown in the
figure. What is the magnitude of the single
resultant force acting on the body and at what angle to the horizontal does it
act? Give your answers correct to one
decimal place.
As specified in the question, we
have three forces we’re interested in. We’ve got 25 newtons acting
vertically upwards, 10 newtons acting to the right horizontally, and five newtons
acting at an angle. The question is asking us to find
some information about the single resultant force acting on the body.
There is a little bit of a problem
though. We have coplanar forces. They’re all acting in one
plane. And we have two forces
perpendicular to one another. But then we have this force acting
at an angle. We need to take into account that
that force acting at an angle will have components that act horizontally and
components that act vertically. So we’re going to split that
five-newton force into its horizontal and vertical components. By doing so, we’ll be able to find
the resultant of the horizontal components and the resultant of the vertical
components. And that’s because the resultant
force of two or more forces is the vector sum of those forces.
To resolve our five-newton force
into its horizontal and vertical components, we’re going to drop in a right-angled
triangle as shown. We’ll call the horizontal component
of this force 𝑥, and the vertical component we’ll call 𝑦. The magnitude of this force is five
newtons. So that’s actually the length of
the hypotenuse in this right-angled triangle. Then we can use trigonometric
convention to label our triangle with respect to the 45-degree angle. The side 𝑥 is the side adjacent to
our included angle. The side labeled 𝑦 is the side
opposite it. And we already saw that we knew the
hypotenuse is five newtons.
To calculate the horizontal
component 𝑥, we’re going to link the adjacent and hypotenuse by using the cosine
ratio. And substituting what we know about
our triangle into this formula, we get cos of 45 degrees equals 𝑥 divided by
five. To solve for 𝑥, we simply multiply
both sides of our equation by five. And we find that 𝑥 is five times
cos of 45 degrees. Now, in fact, we know cos of 45
degrees is root two over two. So 𝑥 is five root two over two,
and in fact that’s in newtons.
We need to repeat this process to
find the value of 𝑦. This time, we use the sine ratio
since that links the opposite side with the hypotenuse. And substituting what we know into
the formula, we get sin of 45 equals 𝑦 over five. Once again, we multiply both sides
of this equation by five. And we find 𝑦 is equal to sin of
45 degrees, and that’s five root two over two newtons.
Now, it’s very interesting that 𝑥
and 𝑦 are the same. And if we go back to our
right-angled triangle, we see that we could’ve saved ourselves a little bit of
time. It’s a right-angled triangle for
which one of the angles is 45 degrees. This means that the third angle,
the angle at the top of our triangle here, must also be 45 degrees. And we, therefore, have an
isosceles triangle, and 𝑥 and 𝑦 must in fact be the same. And so we know both 𝑥 and 𝑦 are
five root two over two.
Our next job is to find the vector
sum of all of our forces. To do so though, we’re going to
need to define a positive direction. As is convention on the 𝑥𝑦-plane,
it makes sense that horizontally to the right would be positive in the 𝑥-direction
and vertically upwards would be positive in the 𝑦-direction.
Let’s begin by considering the sum
of the forces acting horizontally. We have that 10-newton force, and
then we have the horizontal component of our five-newton force. Their sum is simply 10 plus five
root two over two. Notice that since the five-newton
force is acting to the right and up, we can infer that the horizontal component must
be acting in the positive direction. Let’s repeat this process in the
vertical direction. We have the 25-newton force and the
vertical component of the five-newton force. Once again, this vertical component
must be acting upwards, and so it’s positive. And so the sum of the forces in
this direction is 25 plus five root two over two. And whilst not entirely necessary,
we can choose to write this using vector notation. We see the resultant vector 𝐑 in
newtons is 10 plus five root two over two 𝐢 plus 25 plus five root two over two
𝐣.
Now that we know the resultant, we
need to work out the magnitude of that resultant and the angle to the
horizontal. If a direction isn’t given for the
angle, we do assume that we’re taking the angle from the positive horizontal, in
other words, the positive 𝑥-axis. Now, remember, the magnitude of a
vector is the square root of the sum of the squares of its components. So the magnitude of our force is
the square root of 10 plus five root two over two squared plus 25 plus five root two
over two squared. That gives us 31.58, which correct
to one decimal place is 31.6 newtons. And so we have the magnitude of the
resultant. And what about the angle to the
horizontal?
Let’s sketch the resultant vector
𝐑. We now know both the horizontal and
vertical components of this vector. And so we can use right angle
trigonometry to find the angle that the resultant makes with the positive
horizontal. Let’s call that 𝜃. Relative to our included angle, we
know the opposite and the adjacent sides. And so we use the tan ratio. We find that tan of 𝜃 is 25 plus
five root two over two divided by 10 plus five root two over two. And then if we evaluate that
fraction, we get 2.108 and so on. To find the value of 𝜃, we simply
find the inverse or arc tan of both sides of this equation. So 𝜃 is the inverse tan of 2.108
and so on, which is 64.623. Correct to one decimal place,
that’s 64.6. The magnitude of our single
resultant force then is 31.6 newtons, and the angle to the horizontal it makes is
64.6 degrees.
At this stage, it’s really
important to note that whilst the process for finding the magnitude of the resultant
is generally the same, the process for finding the angle it makes with the positive
horizontal axis isn’t. If both the horizontal and the
vertical components of our resultant are positive, we can simply apply the fact that
our angle 𝜃 will be the inverse tan of the 𝐣-component divided by the
𝐢-component. If this is not the case though,
it’s really sensible to draw out a diagram and just see where our resultant force
lies. We will be able to use the tan
ratio, but we might need to do some adding or subtracting from 180.
In our final example, we’ll see how
knowing the resultant can help us to find missing values.
Coplanar forces of magnitudes 𝐹
newtons, eight root three newtons, root three newtons, and nine root three newtons
act on a particle, as shown in the diagram. Given that the magnitude of their
resultant is nine root three newtons, determine the value of 𝐹.
We’re given information about the
magnitude of the resultant of four coplanar forces, where of course the resultant is
the vector sum of those forces. The problem is, one of those forces
is 𝐹, and that’s what we’re trying to find. So what we’re going to do is
resolve each of our forces into perpendicular components. And then we’re going to find the
vector sum of these and then find the magnitude and equate that to nine root
three. So let’s define the direction in
which the 𝐹 force is acting to be the positive horizontal direction. And then perpendicular to this and
upwards is the positive vertical direction.
Let’s begin by resolving our forces
then in a horizontal direction. We know 𝐹 is acting in this
direction. But actually, we need to find the
horizontal component of the eight root three, nine root three, and root three
forces. We add a right-angled triangle to
our eight root three force. We know that the horizontal
component will be acting to the right and the vertical component this will be acting
upwards. And then we can use right angle
trigonometry to see that the adjacent side, and that’s the horizontal component, is
eight root three cos 60 and the opposite side, the vertical component, is eight root
three sin 60.
Let’s repeat this for the root
three force. Since angles on a straight line add
up to 180 degrees, the included angle this time is 60. And then we see that the side
opposite to this, that’s the vertical component acting upwards, is root three sin
60. And the adjacent, which is the
horizontal component acting to the left, is root three cos 60. There’s one more triangle that
we’re interested in. Finally, we have the nine root
three force. We could add a triangle to the left
of this, but there’s an awful lot going on there already. So we add a triangle to the right
as shown.
We know that this angle here is 60
degrees. So the included angle must be
30. And once again, we resolve it into
its horizontal and vertical components. We get nine root three cos of 30
for the vertical component and nine root three sin of 30 for the horizontal
component.
Now that we’ve resolved all of
these forces, let’s find the horizontal sum. We have 𝐹 acting in a positive
direction plus eight root three cos of 60. That’s the component of the eight
root three force that acts in the horizontal direction. And then we subtract root three cos
of 60 since that’s acting to the left, and nine root three sin of 30. By using the fact that both sin of
30 and cos of 60 are one-half, we simplify this to 𝐹 plus eight root three times
one-half minus root three times one-half minus nine root three times one-half. And all that rather complicated
working simplifies to just 𝐹 minus the square root of three.
Let’s do the same in the vertical
direction. We know that we have eight root
three sin 60 acting upwards, so in the positive direction, and root three sin
60. But in the opposite direction, we
have nine root three cos of 30. So we’re going to subtract this
from our sum. This time, we use the fact that sin
of 60 and cos of 30 are root three over two. And so we get eight root three
times root three over two plus root three times root three over two minus nine root
three times root three over two, which is in fact equal to zero. And that’s really useful because we
can actually work out the magnitude of the resultant really easily.
Since our force is only acting in
one direction, the magnitude of our resultant must be 𝐹 minus root three. But we were told that the magnitude
of the resultant is nine root three. So we can set up a simple equation
for 𝐹. 𝐹 minus the square root of three
is equal to nine root three. And we’ll solve for 𝐹 by adding
root three to both sides. And so we see that 𝐹 is 10 root
three or 10 root three newtons.
We’ll now recap the key points from
this lesson. In this video, we learned that
coplanar forces are forces which act in just one plane. And the resultant of a set of
coplanar forces is the vector sum of all of those forces. We also saw that if the forces were
given in vector form, this was really straightforward. But if they’re not, we need to
split them into perpendicular directions, usually 𝑥 and 𝑦. And that allows us to solve
problems involving the resultant. Of course, when we do this, we
mustn’t forget to define a positive direction.
