### Video Transcript

In this video, we will learn how to
find the resultant of two forces acting on one point and how to find the direction
of the resultant. We will begin by defining a force
and exploring its properties.

Force is defined as the effect of
one natural body upon another. Each force is described in terms of
its magnitude, direction, point of action, and line of action. We represent a force using the
notation vector π
as shown.

Letβs now consider the properties
of the magnitude, direction, point of action, and line of action of a force. The magnitude of a force is its
size, which is measured in newtons. By using a directed line segment to
represent the force π
and drawing the line to a suitable scale, we can use the
length of the line to denote the magnitude. This can be written as shown.

The direction of a force is the
direction in which it acts. By once again using a directed line
segment to represent the force π
, we can use the direction of the arrow to show the
direction of the force.

The point of action of a force is
the point at which it is applied.

And finally, the line of action of
a force is a geometric way to represent how the force is applied. It is drawn as a line through the
point of action in the same direction as π
. We can demonstrate this as
follows.

The diagram shows the force π
represented by the directed line segment π΄π΅. The magnitude of the force is
determined by the magnitude of the line segment π΄π΅. The direction of the arrow
corresponds to the direction of π
. The point of action is π΄. And the line of action is indicated
by extending π΄π΅ in the same direction, as shown by the dotted line.

Letβs now consider what happens
when two forces act on a body together with their combined effect, which is known as
the resultant. When two forces π
sub one and π
sub two act on a body at the same point, the combined effect of these two forces is
the same as the effect of a single force called the resultant force. The resultant force vector π is
given by π is equal to π
sub one plus π
sub two. The vector quantity π, which is
equal to π
sub one plus π
sub two, can be represented in two ways as shown. As π
sub one, π
sub two, and π
are three sides of a triangle, we can use either the law of sines or the law of
cosines in the triangle to find the resultant of the two forces, the angles between
the resultant and the forces, or any other unknown.

If we let πΌ be the angle between
forces π
sub one and π
sub two, π sub one be the angle between π and π
sub one,
and π sub two be the angle between π and π
sub two as shown, then using our
knowledge of corresponding and alternate angles, the law of sines gives us πΉ sub
one over sin of π sub two is equal to πΉ sub two over sin of π sub one, which is
equal to π
over sin of 180 degrees minus πΌ, where πΉ sub one, πΉ sub two, and π
are the magnitudes of vectors π
sub one, π
sub two, and π, respectively. We recall that sin of 180 degrees
minus π₯ is equal to sin π₯. And as such, sin of 180 degrees
minus πΌ is equal to sin πΌ. Our equation simplifies as
shown.

Applying the law of cosines in our
triangle, we have π
squared is equal to πΉ sub one squared plus πΉ sub two squared
minus two multiplied by πΉ sub one multiplied by πΉ sub two multiplied by cos of 180
degrees minus πΌ. cos of 180 degrees minus π₯ is
equal to negative cos π₯. And as such, our equation
simplifies as shown.

We now have two equations with many
unknowns that we can simplify further. If we take the square root of both
sides of the second equation and recalling that the magnitude of a vector is
positive, we can obtain an explicit formula for π
, the magnitude of vector π. It is also straightforward to
derive an accompanying formula for the direction of vector π. We will state these results
now.

Let vector π be the resultant of
two forces vector π
sub one and vector π
sub two that act at a single point with
an angle πΌ between them. Then, π
is equal to the square
root of πΉ sub one squared plus πΉ sub two squared plus two πΉ sub one πΉ sub two
multiplied by cos πΌ. And tan π is equal to πΉ sub two
sin πΌ divided by πΉ sub one plus πΉ sub two cos πΌ, where πΉ sub one, πΉ sub two,
and π
are the magnitudes of vectors π
sub one, π
sub two, and π,
respectively. And π is the angle between vector
π and vector π
sub one.

We will now consider two examples,
one where our two forces are perpendicular and one when they are not.

Two perpendicular forces of
magnitudes 88 newtons and 44 newtons act at a point. Their resultant makes an angle
π with the 88-newton force. Find the value of sin π.

In this question, it will be
convenient to assume that one of the forces acts horizontally. We will let this be the
88-newton force and call it vector π
sub one. We will therefore call the
44-newton force vector π
sub two, and this will act vertically. Since the two perpendicular
forces act at a point, we can represent this as shown. The resultant force, vector π,
will act in the direction shown. And we are told that this makes
an angle of π with the 88-newton force π
sub one.

Since vectors have both
direction and magnitude, we can create a triangle of forces. This is a right triangle, and
we can use the properties of right triangles to find the value of sin π as
required. As already mentioned, we know
that the magnitude of vector π
sub one is 88 newtons and the magnitude of π
sub two is 44 newtons. Applying the Pythagorean
theorem, we have π
squared is equal to πΉ sub one squared plus πΉ sub two
squared, where π
, πΉ sub one, and πΉ sub two are the magnitudes of the
corresponding vector forces. This means that π
squared is
equal to 88 squared plus 44 squared. The right-hand side of our
equation simplifies to 9680. We can then square root both
sides. And since π
must be positive,
we have π
is equal to 44 root five. The magnitude of the resultant
force is 44 root five newtons.

Next, we use our knowledge of
right angle trigonometry, often referred to using the acronym SOH CAH TOA. The sine ratio tells us that
sin π is equal to the opposite over the hypotenuse. This means that in this
question, sin π is equal to the magnitude of π
sub two over the magnitude of
the resultant. Substituting in the values we
know, this is equal to 44 over 44 root five. We can then divide the
numerator and denominator by 44, giving us one over root five. Finally, we can rationalize the
denominator by multiplying the numerator and denominator by root five, giving us
sin π is equal to root five over five. This is the sine of the angle
that the resultant makes with the 88-newton force.

Letβs now consider a second example
where the angle between the two forces is not 90 degrees.

The angle between forces vector
π
sub one and vector π
sub two is 112 degrees, and the measure of the angle
between their resultant and vector π
sub two is 56 degrees. If the magnitude of vector π
sub one is 28 newtons, what is the magnitude of vector π
sub two?

Letβs begin by sketching a
diagram to model the situation. We are told that the angle
between two forces π
sub one and π
sub two is 112 degrees. We are also told that the
measure of the angle between the resultant force and π
sub two is 56
degrees. We see that our two forces form
a parallelogram, where the resultant is its diagonal. We can calculate the angle π
between π
sub one and the resultant by subtracting 56 degrees from 112
degrees. This is equal to 56
degrees. We can now add this angle and
its alternate interior angle in our diagram as shown.

Next, we can apply the law of
sines to either one of our triangles within the parallelogram. We have the magnitude of π
sub
one over sin of 56 degrees is equal to the magnitude of π
sub two over the sin
of 56 degrees. This means that the magnitude
of π
sub one is equal to the magnitude of π
sub two. And since we are told in the
question that the magnitude of π
sub one is 28 newtons, then the magnitude of
π
sub two must also be 28 newtons.

We will now summarize the key
points from this video. Force is defined as the effect of
one body upon another. Each force is described in terms of
its magnitude, direction, point of action, and line of action. We represent a force using the
notation vector π
. The resultant vector π of two
forces vector π
sub one and vector π
sub two acting on a body at the same point is
a single force that is given by vector π is equal to vector π
sub one plus vector
π
sub two. The combined effect of vector π
sub one and vector π
sub two is the same as the effect of only vector π. Vectors π
sub one, π
sub two, and
π are three sides of a triangle or two adjacent sides and a diagonal of a
parallelogram.

Applying the law of sines in the
triangle formed by two forces π
sub one and π
sub two and their resultant π gives
πΉ sub one over sin of π sub two equals πΉ sub two over sin of π sub one, which is
equal to π
over sin πΌ, where πΉ sub one, πΉ sub two, and π
are the magnitudes of
vectors π
sub one, π
sub two, and π, respectively. πΌ is the angle between forces π
sub one and π
sub two. π sub one is the angle between the
resultant π and π
sub one. And π sub two is the angle between
π and π
sub two.

Applying the law of cosines in the
same way under the same conditions gives us π
squared is equal to πΉ sub one
squared plus πΉ sub two squared plus two multiplied by πΉ sub one multiplied by πΉ
sub two multiplied by cos πΌ. Square rooting both sides of this
equation, we have π
equals the square root of πΉ sub one squared plus πΉ sub two
squared plus two multiplied by πΉ sub one multiplied by πΉ sub two multiplied by cos
πΌ. And finally, we saw that if π is
the angle between the resultant π and π
sub one, then tan π is equal to πΉ sub
two sin πΌ divided by πΉ sub one plus πΉ sub two cos πΌ.