Video Transcript
The force 𝐅 is the resultant of
the two force vectors shown in the diagram. What is the magnitude of 𝐅 to the
nearest newton?
In this question, we’ve got a
diagram that shows three vectors. This purple vector here represents
a force 𝐅, and we’re told that 𝐅 is the resultant of the other two force
vectors. So that’s the red arrows here,
which are each labeled with a magnitude and a direction. We’re being asked to find the
magnitude of the vector 𝐅. The magnitude of 𝐅 is the length
of this purple arrow. We can see from the diagram that
since 𝐅 is the resultant of the other two vectors that when these two vectors are
drawn tip to tail like this — so that’s the second vector drawn, starting with its
tail at the tip of the first vector — then the resultant vector 𝐅 goes from the
tail of the first vector to the tip of the second vector.
To find the magnitude of 𝐅 or the
length of this arrow, we can notice that the vector 𝐅 forms the hypotenuse of a
right-angled triangle. We can recall that Pythagoras’s
theorem tells us that for a general right-angled triangle with a hypotenuse of
length 𝑐 and other sides of lengths 𝑎 and 𝑏 that 𝑐 squared is equal to 𝑎
squared plus 𝑏 squared. If we take the square root of both
sides of this expression, we have that 𝑐, the length of the hypotenuse, is equal to
the square root of 𝑎 squared plus 𝑏 squared. In the case of the right-angled
triangle that we’ve identified in this diagram, the length of the hypotenuse is the
magnitude of the vector 𝐅. So, if we can work out the lengths
of the other two sides of the triangle, which we’ve labeled as 𝑎 and 𝑏, then
Pythagoras’s theorem tells us that for this triangle the magnitude of 𝐅 is equal to
the square root of 𝑎 squared plus 𝑏 squared.
In order to find the values of 𝑎
and 𝑏, we’re going to need to identify two more right-angled triangles in our
diagram. The first of these, which we’ve
labeled as triangle one, is the right-angled triangle that has the first of the two
red vectors as its hypotenuse. The other triangle, which we’ve
labeled as two, is the right-angled triangle whose hypotenuse is the second of the
two red vectors. For each of these two triangles
that we’ve identified, we’re given the length of the hypotenuse and the value of one
of the angles. Since this information isn’t
particularly clear in the diagram now that we’ve drawn over it, let’s draw out these
two triangles separately from the diagram.
Triangle one looks like this kind
of shape. And looking back at the diagram we
were given, we can see that the leftmost angle in the triangle is 20 degrees, and
the length of the hypotenuse, which is the magnitude of the first red vector, is 70
newtons. So, this is triangle one. Let’s now do the same thing for the
triangle we labeled two. In this triangle, the leftmost
angle is 70 degrees, and the length of the hypotenuse, which is the magnitude of the
second red vector, is 60 newtons. So then, this is triangle two. If we now look back at the
triangles in this diagram, we can see that 𝑎, the length of the horizontal side in
the orange triangle, is equal to the sum of the horizontal side of triangle one plus
the horizontal side of triangle two.
We could imagine shifting triangle
two vertically downward so that the base or horizontal side of triangle two was at
the same vertical height as the horizontal side of triangle one. And we would find that it slotted
perfectly into this space here. Let’s label the horizontal side in
triangle one as 𝑎 one and in triangle two as 𝑎 two. Then, in this diagram, we can
identify the lengths 𝑎 one and 𝑎 two and then can easily see that 𝑎 must be equal
to 𝑎 one plus 𝑎 two. We can do the same thing for the
vertical sides of the triangles one and two. We’ll label these as 𝑏 one and 𝑏
two, respectively. In the same way as before, in this
diagram, we could imagine shifting triangle one horizontally until its vertical side
lined up with the vertical side of triangle two. We can then identify the lengths 𝑏
one and 𝑏 two on this diagram. And we can see that 𝑏 must be
equal to 𝑏 one plus 𝑏 two.
Let’s quickly take a moment to take
stock of the situation so far. We’ve got this equation here, which
came from Pythagoras’s theorem. And it tells us how to calculate
the magnitude of the force 𝐅 if we know the values of the quantities we’ve labeled
as 𝑎 and 𝑏. Then, we’ve also got these two
equations here, which tell us how to calculate 𝑎 and 𝑏 if we know the values of 𝑎
one and 𝑎 two and 𝑏 one and 𝑏 two. So, our approach is going to be to
start by finding the values of 𝑎 one, 𝑎 two, 𝑏 one, and 𝑏 two. Then, we’ll use these values to
work out the quantities 𝑎 and 𝑏. And then finally, we can use the
values of 𝑎 and 𝑏 to work out the magnitude of the force 𝐅.
To find the values of these
quantities 𝑎 one, 𝑎 two, 𝑏 one, and 𝑏 two, we’re going to need to use a couple
of trigonometrical formulas. Let’s consider a general
right-angled triangle, and we’ll suppose that this angle has some value of 𝜃. We’ve labeled the side of the
triangle opposite this angle as capital 𝑂 for opposite, the side adjacent to the
angle as capital 𝐴 for adjacent, and the hypotenuse as capital 𝐻 for
hypotenuse. The two formulas that we’re going
to need are these ones right here. The first one says that cos 𝜃 is
equal to 𝐴 divided by 𝐻. So, the cos of this angle 𝜃 is
equal to the length of the side that’s adjacent to it divided by the length of the
hypotenuse.
The second equation says that sin
𝜃 is equal to 𝑂 divided by 𝐻. So, the sin of this angle 𝜃 is
equal to the length of the side that’s opposite it divided by the length of the
hypotenuse. Looking at the triangles we’ve
labeled as one and two, we can see that the lengths we’re trying to find are the
adjacent and opposite sides on these triangles. This means that we want to
rearrange these two equations in order to make the quantities 𝐴 and 𝑂 the
subject. For both of these two equations,
the process is the same. We simply need to multiply both
sides by the hypotenuse 𝐻. When we do this rearrangement, we
find that the adjacent side 𝐴 is equal to 𝐻 times cos 𝜃 and the opposite side 𝑂
is equal to 𝐻 times sin 𝜃.
Let’s now apply these two equations
to our triangles one and two, starting with triangle one. In this case, the length of the
hypotenuse is 70 newtons, and the angle which corresponds to 𝜃 is 20 degrees. Then, the adjacent side, 𝑎 one,
must be equal to the hypotenuse of 70 newtons multiplied by the cos of the angle of
20 degrees. Similarly, the side opposite the
angle, which is 𝑏 one, must be equal to 70 newtons, the hypotenuse, multiplied by
the sin of the angle 20 degrees. Evaluating the two expressions for
the sides 𝑎 one and 𝑏 one, we find that 𝑎 one is equal to 65.778 newtons and 𝑏
one is equal to 23.941 newtons. In each case, the ellipses are used
to indicate that the answers have further decimal places.
Let’s now clear some space and do
the same thing for triangle two. In this case, the hypotenuse has a
length of 60 newtons and the angle which corresponds to 𝜃 has a value of 70
degrees. So, we have that the adjacent side
𝑎 two is equal to 60 newtons multiplied by the cos of 70 degrees and the opposite
side 𝑏 two is equal to 60 newtons multiplied by the sin of 70 degrees. 𝑎 two works out as 20.521 newtons,
and 𝑏 two works out as 56.382 newtons. Now that we found the values for
the quantities 𝑎 one, 𝑎 two, 𝑏 one, and 𝑏 two, let’s clear ourselves some space
so that we can use those values in these two equations to calculate the values of 𝑎
and 𝑏.
We know that 𝑎 is equal to 𝑎 one
plus 𝑎 two. And substituting in these values
for the quantities 𝑎 one and 𝑎 two and then evaluating the expression that we get,
we find that 𝑎 is equal to 86.299 newtons. In the same way, we know that 𝑏 is
equal to 𝑏 one plus 𝑏 two. And substituting in our values for
𝑏 one and 𝑏 two and then evaluating the expression, we find that 𝑏 is equal to
80.323 newtons. We’re going to clear ourselves some
space one last time so that we can take our values for the quantities 𝑎 and 𝑏 and
substitute them into this equation to calculate the magnitude of the force 𝐅.
Okay, so we’ve got our value for 𝑎
which is the horizontal side of the orange triangle. And we’ve got our value for 𝑏,
which is the vertical side of the same triangle. And we’ve got our equation from
Pythagoras’s theorem, which tells us that the hypotenuse of this triangle, which is
the magnitude of the force 𝐅, is equal to the square root of 𝑎 squared plus 𝑏
squared. If we now substitute in our values
for 𝑎 and 𝑏, we get this expression for the magnitude of 𝐅.
Evaluating this sum under the
square root, we find that the magnitude of 𝐅 is equal to the square root of
13899.30 newtons squared. Taking the square root gives a
result to two decimal places of 117.90 newtons. We’re told in the question to give
our answer to the nearest newton. By rounding this result to the
nearest newton, we get our answer to the question that the magnitude of the force 𝐅
is equal to 118 newtons.
