Video Transcript
Two forces of magnitudes six 𝐹 and
12𝐹 act at a point. The magnitude of their resultant is
nine 𝐹. Find the measure of the angle
between them, giving your answer to the nearest minute.
Let’s begin by imagining what these
two forces might look like. They act at some point 𝑃. So let’s suppose the force of
magnitude six 𝐹 acts in the horizontal direction as shown. The force of magnitude 12𝐹 acts at
the same point, but it must act in a different direction. We might assume it acts
approximately in the direction shown. We’re trying to find the measure of
the angle between these two forces. Let’s call that 𝜃. Now, in order to find the value of
𝜃, we need to use the fact that the magnitude of the resultant of these two forces
is nine 𝐹.
Remember, the resultant of the two
forces is simply their sum. But that doesn’t mean we add six 𝐹
and 12𝐹. These are their magnitudes. So we’re going to instead redraw
our diagram as a triangle of forces. To do so, we begin with the six-𝐹
force as before. We then model the force of
magnitude 12𝐹 so that it starts at the terminal point of the force of magnitude six
𝐹. Since the resultant is the sum of
the forces, the resultant can be modeled as starting at the initial point of the
six-𝐹 force and finishing at the terminal point of the force of magnitude 12
newtons as shown.
We now notice that this is a
non-right triangle and we know the lengths, which are the magnitudes of each force
of all three sides. We can therefore find the measure
of the angle between the six-𝐹 force and the 12𝐹 force in this triangle. Let’s call that 𝛼. We can then use information about
angles in parallel lines to find the value of 𝜃. So how do we find the value of
𝛼? Well, let’s label our triangle as
shown. Since we know all three sides and
we’re trying to find an angle, we can use the law of cosines. We can write that as cos 𝐴 equals
𝑏 squared plus 𝑐 squared minus 𝑎 squared over two 𝑏𝑐. Since capital 𝐴 is angle 𝛼, the
left-hand side is simply cos 𝛼.
Then, we substitute everything we
know about each side in our triangle. On the right-hand side, we then get
12𝐹 squared plus six 𝐹 squared minus nine 𝐹 squared over two times 12𝐹 times six
𝐹. This becomes 144𝐹 squared plus
36𝐹 squared minus 81𝐹 squared over 144𝐹 squared, which simplifies even further to
99𝐹 squared over 144𝐹 squared. And of course, 𝐹 is part of the
magnitude of our force, so we know it cannot be equal to zero. This means we’re able to divide
both the numerator and denominator of our fraction by 𝐹 squared. So the cosine of our angle 𝛼 must
be equal to 99 over 144.
To find the measure of angle 𝛼
then, let’s take the inverse or arccos of both sides. The inverse cosine of 99 over 144
is 46.567 and so on. And of course, that’s in
degrees. Now, we won’t round this just yet
because we’re looking to find 𝜃, not 𝛼. Now, if we go back to our earlier
diagrams and compare them both, we see that angle 𝜃 and angle 𝛼 can be linked by a
pair of parallel lines. In fact, they’re supplementary
angles. This means that 𝜃 plus 𝛼 must be
equal to 180 degrees. So 𝜃 is 180 minus 𝛼. And of course, we’ve calculated
𝛼. So 𝜃 must be equal to 180 minus
that exact value of 46.567 and so on. That gives us 133.432.
Now the question asks us to give
our answer to the nearest minute. To do so, we need to extract the
decimal part of our answer and multiply it by 60. 0.432 and so on multiplied by 60 is
25.95. Now, of course, correct to the
nearest whole number, that will be 26. And so we found the measure of the
angle between our two forces. Correct to the nearest minute, it’s
133 degrees and 26 minutes.
