### Video Transcript

In this video, we will learn how to
find the resultant of a system of parallel coplanar forces and how to locate its
point of action. We recall that multiple forces that
act at a point sum to a resultant force that acts at that point. And the line of action of the
resultant force intersects the lines of action of its components at the point where
the forces act. In this video, however, we will
deal with forces that do not act at the same point, as shown in the diagram.

Letβs begin by considering the
scenario where we have three parallel coplanar forces, π
sub one, π
sub two, and
π
sub three, as shown in the diagram. Finding the resultant force
together with the point at which it acts is slightly more complicated than when the
forces act at one point. There is, however, a step-by-step
process we can apply.

As forces are vectors, they have
direction as well as magnitude. So our first step is to set up a
sign convention. In this example, we will let the
positive direction be vertically upwards. Our next step is to find the sum of
the parallel forces. Since π
sub one and π
sub three
act upwards, they will be positive, whereas π
sub two acts downwards so will
therefore be negative. The resultant force is equal to π
sub one minus π
sub two plus π
sub three.

Letβs now assume that the three
forces have magnitude two newtons, three newtons, and four newtons,
respectively. The resultant force is therefore
equal to two minus three plus four, which is equal to three newtons. As this is positive, we know that
the resultant force acts vertically upwards with magnitude three newtons.

We now need to consider where along
our line this resultant force acts. In order to do this, we model the
line as a light thin rod and then take moments about any point on the line. This is the third step of our
process. Typically, when taking moments, we
consider counterclockwise moments to be positive and clockwise moments to be
negative. Whilst we can take moments about
any point, typically, we choose a point where one of the forces acts. In this case, weβll take moments
about the point where the force π
sub one acts.

As already mentioned, we do not
know where the resultant three-newton force acts. However, we will add it to our
diagram a perpendicular distance of π₯ meters from where the force π
sub one
acts. We know that we can calculate the
moment of a force by multiplying the magnitude of the force by the perpendicular
distance from the force to the point at which we are taking moments. This is usually written as π is
equal to π
multiplied by π.

Returning to our diagram and the
general notation π
sub one, π
sub two, and π
sub three, we can add the distances
π sub two and π sub three. These are the perpendicular
distances of π
sub two and π
sub three from the point at which we are taking
moments. As the π
sub one force is acting
at the point at which we are taking moments, this will have a moment equal to
zero. The resultant force is acting in a
counterclockwise direction about the point at which we are taking moments. It will therefore have a positive
moment equal to π
multiplied by π₯. The π
two newton force acts in a
clockwise direction, so it will have a negative moment. This is equal to negative π
sub
two multiplied by π sub two. Finally, the π
sub three force is
acting in a counterclockwise direction so will have a positive moment. This is equal to π
sub three
multiplied by π sub three.

If π sub two was equal to two
meters and π sub three four meters, we can substitute all these values into our
equation. This gives us three π₯ is equal to
negative three multiplied by two plus four multiplied by four. The right-hand side simplifies to
negative six plus 16, and this is equal to 10. We can then divide through by three
such that π₯ is equal to 10 over three or ten-thirds. The resultant force π
acts a
perpendicular distance of 10 over three meters from the π
sub one force.

This leads us to a general rule
that we can use to calculate the perpendicular distance of the resultant from a
point at which we are taking moments. The distance π₯ is equal to the sum
of the moments divided by the resultant force. We will now look at some examples
where we need to follow this four-step process.

Two parallel forces have
magnitudes of 10 newtons and 20 newtons. The distance between their
lines of action is 30 centimeters. If the two forces are acting in
the same direction, find their resultant π
and the distance π₯ between its line
of action and point π΄.

We begin by noticing from the
question that we are dealing with two parallel coplanar forces. Both of these forces act in the
positive π¦-direction and have magnitudes of 10 and 20 newtons. The distance between the two
forces is 30 centimeters. We are asked to find the
resultant π
and the distance π₯ between the resultantβs line of action and
point π΄.

We know that when dealing with
parallel coplanar forces, the resultant force π
is equal to the sum of the
other forces. Since both forces are acting in
the positive π¦-direction, we have π
is equal to 10 plus 20. This is equal to 30. The resultant force π
is equal
to 30 newtons and acts vertically upwards.

We donβt know at present where
this resultant force acts. Letβs assume it acts a
perpendicular distance of π₯ centimeters from π΄. By considering moments, we
recall that this distance π₯ is equal to the sum of the moments divided by the
resultant force π
. We know that the moment of a
force is equal to the magnitude of the force multiplied by the perpendicular
distance to the point at which we are taking moments. We will consider moments in the
counterclockwise direction to be positive and take moments about point π΄.

The moment of the 10-newton
force will therefore be equal to zero, as this acts at that point. The only other force we need to
consider is the 20-newton force. This will have a moment of 20
newtons, its magnitude, multiplied by 30 centimeters, its distance from point
π΄. 20 multiplied by 30 is equal to
600. Therefore, the sum of the
moments is equal to 600 newton centimeters. Substituting this into our
formula, we have π₯ is equal to 600 divided by 30. This is the same as 60 divided
by three. π₯ is therefore equal to 20
centimeters. The distance between the line
of action of the resultant force and point π΄ is therefore equal to 20
centimeters.

We will now consider an example
where the parallel forces act in opposite directions.

Two parallel forces have
magnitudes of 24 newtons and 60 newtons as shown in the figure. The distance between their
lines of action is 90 centimeters. Given that the two forces are
acting in opposite directions, determine their resultant π
and the distance π₯
between its line of action and point π΄.

In this question, we have two
parallel coplanar forces acting in opposite directions. They have magnitudes 24 and 60
newtons, where the 24-newton force acts in the positive direction and the
60-newton force acts in the negative direction. We are also told that the
distance between their lines of action is 90 centimeters. We have been asked to calculate
the resultant force π
together with the distance π₯ between its line of action
and point π΄.

We know that the resultant
force is equal to the sum of the other forces. Since the positive direction is
vertically upwards, this is equal to 24 plus negative 60, which is equal to
negative 36. The resultant force π
is equal
to negative 36 newtons. This means that it acts in the
downward direction with magnitude 36.

We donβt know at present where
this resultant force acts. But we can calculate the
distance π₯ between its line of action and point π΄ using the formula π₯ is
equal to the sum of the moments divided by the resultant force π
, where the
moment of any force can be calculated by multiplying the magnitude of that force
by the perpendicular distance to the point at which we are taking moments.

In this question, weβll take
moments about point π΄, where moments acting in the counterclockwise direction
are positive. The 24-newton force acts at
point π΄, which means it is a distance of zero centimeters from point π΄. It will therefore have a moment
equal to zero. This means that the only moment
we need to consider is that of the 60-newton force. This acts in a clockwise
direction about point π΄. Therefore, the moment is equal
to negative 60 multiplied by 90, where 90 centimeters is the perpendicular
distance of this force from point π΄. 60 multiplied by 90 is
5400. Therefore, negative 60
multiplied by 90 is negative 5400.

We can therefore calculate π₯
by dividing negative 5400 by negative 36. Dividing a negative number by a
negative number gives a positive answer. Therefore, π₯ equals 150. The distance between the line
of action of the resultant force and point π΄ is 150 centimeters.

We can see from this answer an
interesting point about this question. The line of action of the
resultant force does not lie between the line of action of our other forces. In practice, we could consider
a light thin rod π΄πΆ of length 150 centimeters. If forces of magnitude 24
newtons and 60 newtons are acting upon it as shown, then we would require a
vertical force of magnitude 36 newtons at point πΆ in order to maintain the
equilibrium.

We will now look at one final
example where the parallel coplanar forces are joined by a line that is not
perpendicular to their lines of action.

In the figure below, π
sub one
and π
sub two are two parallel forces measured in newtons, where π
is their
resultant. If π
is equal to 30 newtons,
π΄π΅ equals 36 centimeters, and π΅πΆ equals 24 centimeters, determine the
magnitude of π
sub one and π
sub two.

In this question, we have two
parallel coplanar forces, π
sub one and π
sub two, acting in opposite
directions. We are also given the resultant
force π
, which is equal to 30 newtons. The distance from point π΄ to
π΅ is 36 centimeters, and the distance from π΅ to πΆ is 24 centimeters. The lines of action of our
forces π
sub one, π
sub two, and π
are not perpendicular to the line segment
π΄πΆ. However, as our three forces
are parallel, we can add the angle π to our diagram as shown.

We can calculate the
perpendicular components of these forces using our knowledge of right angle
trigonometry. These are equal to π
sub one
sin π, π
sub two sin π, and π
sin π. These will be useful when we
come to take moments, as the moment of a force is equal to the magnitude of
force multiplied by the perpendicular distance to the point at which we are
taking moments. Whilst we can take moments
about any point on our line, in this question, we will do so about point π΄.

We will consider moments acting
in the counterclockwise direction to be positive and those acting in the
clockwise direction to be negative. This means that the moment π
sub one of the force π
sub one acts in the positive direction and is equal to
π
sub one sin π multiplied by 36. We can repeat this process for
π sub two, which is the moment of the force π
sub two. As this acts in the negative
direction, this is equal to negative π
sub two sin π multiplied by 60.

Our expressions for π sub one
and π sub two can be simplified as shown. We know that the distance π₯
from the line of action of the resultant force to the point at which we are
taking moments is equal to the sum of the moments divided by π
. In this case, we are taking
moments about the point where the resultant acts. Therefore, π₯ is equal to
zero. This means that the sum of our
two moments must equal zero. 36 multiplied by π
sub one sin
π plus negative 60 multiplied by π
sub two sin π equals zero. Since sin π cannot be equal to
zero, we can divide through by this. We can also divide through by
12 such that three π
sub one minus five π
sub two equals zero. As there are two unknowns here,
we will call this equation one.

We will now consider the
resultant force and the fact that this is equal to the sum of the other
forces. Going back to our initial
diagram, if we let the positive direction be vertically upwards, we have π
is
equal to π
sub one plus negative π
sub two. Since π
is equal to 30
newtons, we have 30 is equal to π
sub one minus π
sub two. Adding π
sub two to both sides
of this equation, we have π
sub one is equal to 30 plus π
sub two. We will call this equation
two. And we now have a pair of
simultaneous equations that we can solve by substitution.

One way of doing this is to
substitute the expression for π
sub one in equation two into equation one. This gives us three multiplied
by 30 plus π
sub two minus five π
sub two is equal to zero. Distributing the parentheses
gives us 90 plus three π
sub two. The left-hand side then
simplifies to 90 minus two π
sub two. By adding two π
sub two to
both sides and then dividing through by two, we have π
sub two is equal to
45. Substituting this value back
into equation two gives us π
sub one is equal to 30 plus 45, which is equal to
75. The magnitude of the two forces
π
sub one and π
sub two are 75 newtons and 45 newtons, respectively.

We will now summarize the key
points from this video. We saw in this video that parallel
coplanar forces can be combined into a resultant force π
such that π
is equal to
the sum of π
. The point of action of π
is the
point at which the moments about the point due to the forces sum to zero. This means that we can calculate
the distance π₯ from the line of action of the resultant force to any point on the
line by finding the sum of the moments and dividing it by the resultant force
π
.