Video Transcript
In this video, we will learn how to
find the resultant of a group of forces acting at a point.
Suppose that multiple forces act at
a point, such as in the figure. We call the net force exerted by
the combination of all the forces the resultant force. In this case, it is the vector
𝐑. To calculate the resultant of any
number of forces, we can add the forces together tip to tail. That is, we move the starting
point, or tail, of 𝐅 two to the finishing point, or tip, of 𝐅 one. We move the tail of 𝐅 three to the
tip of 𝐅 two, and so on for however many forces we have. The resultant force is given by the
vector from the starting point to the tip of the final force, 𝐅 three. 𝐑 is the vector sum of the forces
𝐅 one, 𝐅 two, and 𝐅 three.
Another way to find the resultant
is to add the perpendicular components of each force together. Suppose we have a force 𝐅 one that
can be decomposed into the horizontal and vertical components 𝐅 one 𝑥 and 𝐅 one
𝑦. Similarly, we have a force 𝐅 two
that can be decomposed into 𝐅 two 𝑥 and 𝐅 two 𝑦. Then, we can find the resultant of
these two forces by adding the components as shown. 𝐑 is the vector sum of 𝐅 one and
𝐅 two. And the horizontal component of 𝐑,
𝐑 𝑥, is the sum of the horizontal components 𝐅 one 𝑥 and 𝐅 two 𝑥. The vertical component of 𝐑, 𝐑
𝑦, is the sum of the vertical components 𝐅 one 𝑦 and 𝐅 two 𝑦.
The horizontal component, 𝐅 𝑥,
and the vertical component, 𝐅 𝑦, of a force 𝐅 can be found from the inclination
angle of 𝐅. The horizontal component 𝐅 𝑥 is
given by the magnitude of the force 𝐅 multiplied by the cosine of the adjacent
angle 𝜃 or by the sine of the opposite angle 𝜙. The vertical component 𝐅 𝑦 is
given by 𝐅 multiplied by sin 𝜃 or 𝐅 multiplied by cos 𝜙. The magnitude of the force 𝐅 is
given by the square root of the sum of the components squared. From this, we can also derive that
the tan of 𝜃 is equal to the vertical component 𝐅 𝑦 over the horizontal component
𝐅 𝑥. And it follows that 𝜃 is equal to
the inverse tan of 𝐅 𝑦 over 𝐅 𝑥.
Let us now look at an example which
includes multiple forces acting at a point.
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.
To begin, the five-newton force can
be decomposed into its horizontal and vertical components. We can then add these components to
the other horizontal and vertical forces to find the resultant force. The horizontal component is equal
to the magnitude, five, multiplied by the cosine of the adjacent angle, 45
degrees. The vertical component is equal to
five multiplied by the sine of the adjacent angle, 45 degrees. Adding together all horizontal
components, we get the horizontal component of the resultant force, 𝐅 𝑥, equals 10
plus five times cos 45 degrees.
Likewise, the vertical component of
the resultant force, 𝐅 𝑦, is equal to 25 plus five times sin 45 degrees. Both cos of 45 degrees and sin of
45 degrees are equal to the square root of two over two. So these become 10 plus five times
the square root of two over two and 25 plus five times the square root of two over
two, respectively.
Now recall that the magnitude of
any force in two dimensions is given by the positive square root of the sum of its
horizontal and vertical components squared. The inclination angle 𝜃 is given
by the inverse tangent of the vertical component, 𝐅 𝑦, over the horizontal
component, 𝐅 𝑥. Therefore, the magnitude of the
resultant force, 𝐅 𝑅, is given by the square root of 10 plus five times the square
root of two over two squared plus 25 plus five times the square root of two over two
squared. Performing this calculation, we
obtain the magnitude of the resultant force to one decimal place is 31.6
newtons.
The angle of inclination 𝜃 is
equal to the inverse tan of 25 plus five times the square root of two over two
divided by 10 plus five times the square root of two over two. Performing this calculation gives
us the angle the resultant force makes with the horizontal is 64.6 degrees to one
decimal place.
Note, there is no limit to the
number of forces that can act at a point. And forces that act at a point can
be resolved into perpendicular components.
Let’s now look at an example of six
forces acting at a point.
The diagram shows a regular
hexagon, 𝐴𝐵𝐶𝐷𝐸𝐹, whose diagonals intersect at point 𝑀. The six forces shown acting at 𝑀
are measured in newtons. Find 𝑅, the magnitude of their
resultant, and 𝜃, the angle between their resultant and the positive 𝑥-axis. Round your value of 𝜃 to the
nearest minute if necessary.
The regular hexagon consists of six
congruent equilateral triangles, which have an internal angle of 60 degrees. Consider the force acting along the
line 𝑀𝐹. The horizontal component of the
force will be given by the magnitude, 60 newtons, multiplied by the cosine of the
angle the force makes with the positive 𝑥-axis, 60 degrees. Similarly, the vertical component
of the force is equal to the magnitude, 60 newtons, multiplied by the sine of the
angle the force makes with the positive 𝑥-axis, 60 degrees.
For the rest of the forces, the
same formula will apply, but the angle will change. The angles the forces in the
directions of 𝐸, 𝐷, 𝐶, 𝐵, and 𝐴 make with the positive 𝑥-axis are each 60
degrees greater than the last: 120 degrees, 180 degrees, 240 degrees, 300 degrees,
and 360 degrees, respectively.
We can add these horizontal
components together to give the horizontal component of the resultant force as
shown. We can do the same for the vertical
component of the force, this time taking the sine of the angles with the positive
𝑥-axis, as shown. These trigonometric components
evaluate to fairly simple values, which we can substitute in.
We now perform some simplification
on the horizontal component of the force 𝐅 𝑥. Performing this calculation, we
find 𝐅 𝑥 is equal to five. We can do the same with the
vertical components, for which the trigonometric components are also quite
simple. Again, we perform some
simplification steps. Here we find we can take out a
common factor of the square root of three to simplify our answer. We then find that 𝐅 𝑦 is equal to
nine times the square root of three. We will leave this in surd form to
calculate the resultant force.
Now that we have the horizontal and
vertical components of the resultant force, we can find its magnitude by taking the
positive square root of the sum of these components squared. And we can find the inclination
angle of the force by taking the inverse tangent of the vertical component over the
horizontal component. For the magnitude of the force,
this gives us the square root of five squared plus nine times the square root of
three squared. This simplifies to two times the
square root of 63 newtons.
For the inclination angle 𝜃, we
have the inverse tan of nine times the square root of three over five. This is equal to approximately
72.21 degrees, which to the nearest minute is equal to 72 degrees and 13
minutes.
Forces can also be expressed in
terms of their perpendicular components. In a 2D Cartesian coordinate
system, this is done with scalar multiples of the unit vectors 𝐢 and 𝐣, which act
in the 𝑥- and 𝑦-directions, respectively.
Let’s now take a look at an example
of determining the resultant of multiple forces expressed in this way.
The resultant of forces 𝐅 one
equals minus four 𝐢 plus two 𝐣 newtons, 𝐅 two equals five 𝐢 minus seven 𝐣
newtons, and 𝐅 three equals two 𝐢 plus nine 𝐣 newtons makes an angle of 𝜃 with
the positive 𝑥-axis. Determine 𝑅, the magnitude of the
resultant, and find the value of tan 𝜃.
Let us start with a force diagram
of 𝐅 one, 𝐅 two, and 𝐅 three acting from a point. We can determine the horizontal and
vertical components of the resultant by summing the horizontal and vertical
components of the individual forces. The horizontal components of 𝐅
one, 𝐅 two, and 𝐅 three are given by the question as negative four, five, and two
newtons, respectively. Likewise, the vertical components
are given by two, negative seven, and nine newtons, respectively.
Therefore, the horizontal component
of the resultant, 𝑅 𝑥, is given by negative four plus five plus two, which is
equal to three newtons. Similarly, the vertical component
of the resultant, 𝑅 𝑦, is given by two minus seven plus nine, which is equal to
four newtons. This gives us that the resultant
force 𝑅 is equal to three 𝐢 plus four 𝐣 newtons.
We can see a visual representation
of this to scale on the diagram and note that seems correct. The magnitude of the resultant 𝑅
is given by the square root of the sum of the horizontal and vertical components, 𝑅
𝑥 and 𝑅 𝑦, squared. In this case, we have the square
root of three squared plus four squared. This simplifies to the square root
of 25, which is equal to five newtons.
The tangent of the angle that the
resultant makes with the positive 𝑥-axis is given by the vertical component, 𝑅 𝑦,
over the horizontal component, 𝑅 𝑥. In this case, therefore, tan 𝜃 is
equal to four over three, which is four-thirds.
Since we have the precise
components of these forces and we can draw them to scale, we can also find the
resultant using the tip-to-tail method. We start by drawing 𝐅 one. We then draw 𝐅 two starting from
the tip of 𝐅 one. Finally, we draw 𝐅 three starting
from the tip of 𝐅 two.
The resultant force is given by the
vector from the original starting point to the tip of the final force, 𝐅 three. On a scale diagram, we could
measure the 𝑥- and 𝑦-components of 𝑅 to get the resultant force, which is equal
to three 𝐢 plus four 𝐣 newtons. We note that this agrees with our
previous result. We would calculate the magnitude of
this vector using exactly the same method as we did previously. And we would find this is equal to
five newtons. Likewise, we’d reach the same
answer for tan 𝜃, which we’d calculate by taking the vertical component over the
horizontal component. And this is four over three.
Let’s finish this video by
recapping some key points. Multiple forces can be summed by
summing the perpendicular components of the forces and determining the resultant of
these components. If 𝐅 𝑥 is the horizontal
component of a force 𝐅, the magnitude of 𝐅 𝑥 is given by the magnitude of 𝐅
multiplied by the cosine of the angle between 𝐅 and 𝐅 𝑥. We call this angle 𝜃.
Similarly, if 𝐅 𝑦 is the vertical
component of the force 𝐅, which is perpendicular to 𝐅 𝑥, this is given by 𝐅
times sin 𝜃. The magnitude of the resultant of
the perpendicular components of a force is given by 𝐅 equals the square root of 𝐅
𝑥 squared plus 𝐅 𝑦 squared. If a force 𝐅 𝐴 is given by 𝑥 𝑎
times 𝐢 plus 𝑦 𝑎 times 𝐣 plus 𝑧 𝑎 times 𝐤 and a force 𝐅 𝐵 is given by 𝑥 𝑏
times 𝐢 plus 𝑦 𝑏 times 𝐣 plus 𝑧 𝑏 times 𝐤, the resultant of these forces is
given by 𝐅 𝐴 plus 𝐅 𝐵. This is equal to 𝑥 𝑎 plus 𝑥 𝑏
times 𝐢 plus 𝑦 𝑎 plus 𝑦 𝑏 times 𝐣 plus 𝑧 𝑎 plus 𝑧 𝑏 times 𝐤.
