Video Transcript
In this lesson, we’ll learn how to
solve problems about the equilibrium of a particle under the action of three forces
using the triangle of forces method. By this stage, you’ll probably have
worked with systems of forces fairly extensively using the Pythagorean theorem and
right-angled trigonometry to find the magnitude and direction of these forces and
also to split them into their component parts. You might recall that a particle is
in equilibrium if the vector sum of these forces, in other words, their resultant,
is equal to zero. We’re now interested in what
happens when there are exactly three forces acting at a point.
The key point to remember here is
that when three coplanar forces, in other words, three forces that lie in the same
two-dimensional plane, acting at a point are in equilibrium, they can be represented
in magnitude and direction by the adjacent sides of a triangle taken in order. With that process complete, we can
then use facts about triangles such as trigonometry and similar triangles to solve
problems involving these forces. For instance, let’s imagine we have
three forces 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three acting at a point as
shown. We can represent these
alternatively using a triangle, remembering to take them in order and consider the
direction of each force. We begin by adding the force 𝐹 sub
one as the first side in our triangle.
Then we move on to force 𝐹 sub
two. This begins at the terminal or
endpoint of 𝐹 sub one and travels in the same direction. Finally, we repeat this process
with the third force, 𝐹 sub three. This time, that begins at the
terminal point of 𝐹 sub two and travels in the same direction as in our first
diagram. And there we have it, a triangle of
forces. We’re now going to see how we can
solve problems using this triangle of forces method.
Three coplanar forces 𝐹 sub one,
𝐹 sub two, and 𝐹 sub three are acting on a body in equilibrium. Their triangle of forces forms a
right triangle as shown. Given that 𝐹 sub one is equal to
five newtons and 𝐹 sub two is equal to 13 newtons, find the magnitude of 𝐹 sub
three.
Remember, when three coplanar
forces acting at a point are in equilibrium, they can be represented in magnitude
and direction by the adjacent sides of a triangle taken in order. We actually have the triangle of
forces drawn for us, and we know the magnitude of two of the forces. 𝐹 sub one is equal to five newtons
and 𝐹 sub two is equal to 13 newtons. This triangle now represents the
relative magnitude of each of our forces. And since it forms a right
triangle, we can find the magnitude of the third force by using the Pythagorean
theorem. This tells us that, in a right
triangle, the sum of the squares of the two shorter sides is equal to the square of
the hypotenuse. If we let the hypotenuse be equal
to 𝑐, then we say that 𝑎 squared plus 𝑏 squared equals 𝑐 squared.
In this case, the longest side in
our triangle is the side represented by the 13-newton force. And so, using the magnitudes we’ve
been given and letting the magnitude of 𝐹 sub three be equal to 𝑏 newtons, we can
say that five squared plus 𝑏 squared equals 13 squared. That is, 25 plus 𝑏 squared equals
169. And subtracting 25 from both sides
of this equation, we find 𝑏 squared is equal to 144. To solve for 𝑏, we simply need to
find the square root of both sides of this equation. Now, usually we would find both the
positive and negative square root of 144. But since this represents a
magnitude, we know it absolutely must be positive. And so 𝑏 is equal to the square
root of 144, which is 12. Given that 𝐹 sub one is five
newtons and 𝐹 sub two is 13 newtons then, we can say the magnitude of 𝐹 sub three
is 12 newtons.
In our next example, we’ll look at
how to take three coplanar forces acting at a point and form a triangle of
forces.
In the figure, three forces of
magnitudes 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three newtons meet at a point. The lines of action of the forces
are parallel to the sides of the right triangle. Given that the system is in
equilibrium, find the ratio of 𝐹 sub one to 𝐹 sub two to 𝐹 sub three.
We know that when three coplanar
forces acting at a point are in equilibrium, they can be represented in magnitude
and direction by the adjacent sides of a triangle taken in order. So we’re going to begin by
representing the three forces in our question using a triangle. We’re going to take these in order,
so let’s begin with force 𝐹 sub one. Then, 𝐹 sub two is perpendicular
to 𝐹 sub one. So we can add that force to our
diagram, noting that we must start at the terminal point of 𝐹 sub one. Then, we add 𝐹 sub three starting
at the terminal point of 𝐹 sub two to complete our triangle. This is a right triangle since we
said that 𝐹 sub one and 𝐹 sub two are perpendicular to one another.
We might also notice that the force
𝐹 sub one is parallel to the side in our original triangle measuring 87
centimeters. 𝐹 sub two is colinear to the side
measuring 208.8 centimeters. And then there’s a shared side
represented by this 𝐹 sub three force. Since this is the case, we can say
that the two triangles, that is, the force triangle and the one whose dimensions we
know, must be similar. They’re proportional to one
another. We can therefore say that the
magnitudes of the forces in our triangle of forces will be directly proportional to
the lengths of the respective sides in that original triangle.
And so to find the ratio of 𝐹 sub
one to 𝐹 sub two to 𝐹 sub three, we’re going to find the ratio of the lengths of
the sides in this triangle. Let’s find the length of the third
side then. We’ll label it 𝑥 centimeters. Since this is a right triangle, we
can use the Pythagorean theorem to find the length of 𝑥. For a right triangle whose longest
side is 𝑐 units, the Pythagorean theorem says that 𝑎 squared plus 𝑏 squared
equals 𝑐 squared. In this case, our hypotenuse is 𝑥
centimeters. So the Pythagorean theorem gives us
87 squared plus 208.8 squared equals 𝑥 squared. Evaluating the left-hand side of
this equation and we find that that’s equivalent to 51166.44. To find the value of 𝑥 then, we
find the square root of both sides. The square root of 51166.44 is
226.2. And so the length of the third side
in our triangle is 226.2 centimeters.
Remember though, we’re trying to
find the ratio of the forces 𝐹 sub one to 𝐹 sub two to 𝐹 sub three. And we said that that will be the
same as the ratio of the relevant sides. Listing these in the relevant order
and we find the ratio of 𝐹 sub one to 𝐹 sub two to 𝐹 sub three to be equivalent
to 87 to 208.8 to 226.2. Dividing each of these numbers by a
rather unusual shared factor, that’s 17.4, and we get five to 12 to 13. Alternatively, if we had calculated
87 divided by 226.2 and 208.8 divided by 226.2, we would have found five thirteenths
and twelve thirteenths, respectively. The ratio of 𝐹 sub one to 𝐹 sub
two to 𝐹 sub three is five to 12 to 13.
In this example, we had some
diagram to go from. In our next example, we’ll need to
draw the triangle of forces completely.
A body is under the effect of three
forces of magnitudes 𝐹 sub one, 𝐹 sub two, and 36 newtons, acting in the
directions of line segments 𝐴𝐵, 𝐵𝐶, and 𝐴𝐶, respectively, where triangle
𝐴𝐵𝐶 is a triangle such that 𝐴𝐵 equals four centimeters, 𝐵𝐶 equals six
centimeters, and 𝐴𝐶 equals six centimeters. Given that the system is in
equilibrium, find 𝐹 sub one and 𝐹 sub two.
We know that when three coplanar
forces acting at a point are in equilibrium, they can be represented in magnitude
and direction by the adjacent sides of a triangle taken in order. So we’re going to represent the
three forces given using a triangle. But we’re told that they act in the
directions of the various sides of triangle 𝐴𝐵𝐶. So we’ll sketch triangle 𝐴𝐵𝐶
first. Triangle 𝐴𝐵𝐶 looks a little
something like this, and we notice that the sides 𝐴𝐶 and 𝐵𝐶 are both six
centimeters in length. So it’s actually an isosceles
triangle.
We’ll now use this triangle to
sketch a triangle of forces representing 𝐹 sub one, 𝐹 sub two, and the 36-newton
force. The force with magnitude 𝐹 sub one
newtons acts in the direction of line segment 𝐴𝐵. Then the force with magnitude 𝐹
sub two acts in the direction of line segment 𝐵𝐶. Notice that the force with
magnitude 𝐹 sub two begins at the terminal point of our previous force. And so we have to begin our third
force at the terminal point of 𝐹 sub two. But we were told that this
36-newton force acts in the direction of the line segment 𝐴𝐶, not the line segment
𝐶𝐴. However, since we know that the
magnitude of this force is 36 newtons, we can label it as shown. If we were considering the
direction of the force, we would need to consider that this would be the negative
direction of our original force. But for magnitudes which just
represent size, this is absolutely fine.
We’re now ready to compare our
triangles. Since each of our forces acts in
the same direction as each side in our triangle 𝐴𝐵𝐶, the two triangles must in
fact be similar. And so we can say that the
magnitudes of each of our forces must be directly proportional to the lengths of the
sides in triangle 𝐴𝐵𝐶. So we can find force 𝐹 sub two
really easily. We know that the sides 𝐴𝐶 and
𝐵𝐶 are equal in length. So this force and this force must
be equal in magnitude. And so 𝐹 sub two must be equal to
36 newtons. And then we have two different ways
that we can calculate the magnitude 𝐹 sub one.
One way is to say that the ratio of
the line segment 𝐴𝐵 to the ratio of line segment 𝐴𝐶 will be equal to the ratio
of the magnitude 𝐹 sub one to the magnitude 36 newtons. In other words, four divided by six
will give us the same outcome as 𝐹 sub one divided by 36. And whilst we could simplify the
fraction four-sixths, it doesn’t make a lot of sense to do this because we’re going
to multiply both sides of this equation by 36. Then we spot that 36 and six have a
common factor of six. So 𝐹 sub one will be equal to six
times four over one, which is simply equal to 24. And so 𝐹 sub one is 24
newtons. It’s worth noting at this point
that we could’ve used scale factor to calculate the value of 𝐹 sub one.
Since the two triangles are
similar, we can deduce that one is an enlargement or a dilation of the other. And thus, the scale factor for
enlargement would be 36, that’s one of the dimensions on our force triangle, divided
by six, the corresponding dimension on triangle 𝐴𝐵𝐶. 36 divided by six is six. And so we can transform any
measurement on our triangle 𝐴𝐵𝐶 onto the dimensions of our force triangle by
multiplying by six. This means 𝐹 sub one would be
equal to four times six, which is once again 24. 𝐹 sub one is 24 newtons and 𝐹 sub
two is 36 newtons.
In our final example, we’ll look at
how to use a similar technique to solve a problem involving tension.
A uniform rod of length 50
centimeters and weight 143 newtons is freely suspended at its ends from the ceiling
by means of two perpendicular strings attached to the same point on the ceiling. Given that the length of one of the
strings is 30 centimeters, determine the tension in each string.
We’re going to begin by drawing a
free-body diagram of this scenario. Let’s define the ends of our rod to
be 𝐴 and 𝐵. And so we have our rod suspended
from two perpendicular strings, one of which measures 30 centimeters. Now, in fact, we can work out the
measurement of the second piece of string. It’s 40 centimeters. And we can calculate that using the
Pythagorean theorem or just recognizing that we have a multiple of a Pythagorean
triple — three, four, five. Since the rod is uniform, we can
say that the downwards force of its weight acts at a point exactly halfway along the
rod. Now, since the rod exerts a
downwards force on the pieces of string, there will be an opposite reaction force of
tension. Let’s call that 𝑇 sub 𝐴 for the
tensional force in the first bit of string and 𝑇 sub 𝐵 for the tensional force in
the second.
We have our free-body diagram, but
there’s still an awful lot going on here. So how do we simplify it
further? Well, we know that when three
coplanar forces acting at a point are in equilibrium, they can be represented in
magnitude and direction by the adjacent sides of a triangle taken in order, since
the systems in equilibrium will lay the force vectors end to end to make a
triangle. 𝑇 sub 𝐴 and 𝑇 sub 𝐵 will be
perpendicular to one another, since the pieces of string along which they run are
also perpendicular to one another. But in fact, with a little bit of
inspection, we can see that these are similar triangles. We’ll define the angle between the
rod and the 30-centimeter piece of string to be 𝑥 degrees. This is equal to the angle between
the 143-newton force and the tensional force at 𝐵.
This isn’t hugely intuitive, but we
can convince ourselves that this is true by adding a line parallel to the 143-newton
force on our first diagram and then using the fact that angles in a triangle sum to
180. Since the angles in our force
triangle and the angles in the triangle represented by the lengths of each item are
equal, we know these triangles are similar. And so we can use scale factor or
ratios to find the magnitudes of 𝑇 sub 𝐴 and 𝑇 sub 𝐵. Let’s consider the ratio of 𝑇 sub
𝐴 to 143. This must be equal to the ratio of
the 40-centimeter length to the 50-centimeter length. And we chose these pairs of sides
in both our diagrams because we’re interested in the pairs of sides that are
adjacent to the angle measuring 90 minus 𝑥 degrees.
Multiplying both sides of this
equation by 143 and we find that 𝑇 sub 𝐴 is 40 over 50 or four-fifths times 143,
which is 114.4. So the magnitude of 𝑇 sub 𝐴 is
114.4 newtons. In a similar way, the ratio of 𝑇
sub 𝐵 to 143 newtons will be equal to the ratio of the 30-centimeter length to the
50-centimeter length. To solve for 𝑇 sub 𝐵 once again,
we multiply by 143, which gives us 85.8 or 85.8 newtons. The tensional forces in each string
are 85.8 newtons and 114.4 newtons.
We’ll now recap the key points from
this lesson. We’ve seen that when three coplanar
forces acting at a point are in equilibrium, they can be represented in magnitude
and direction by the adjacent sides of a triangle taken in order. Once we have this triangle of
forces, we can solve problems by using similarity, the Pythagorean theorem, and even
trigonometry.
