Video Transcript
The resultant of forces 𝐅 sub one
equals negative four 𝐢 plus two 𝐣 newtons, 𝐅 sub two equals five 𝐢 minus seven
𝐣 newtons, and 𝐅 sub three equals two 𝐢 plus nine 𝐣 newtons makes an angle 𝜃
with the positive 𝑥-axis. Determine 𝑅, the magnitude of the
resultant, and the value of tan of 𝜃.
We’re given three vector forces in
this question. These vector forces are given in
terms of 𝐢- and 𝐣-components. 𝐢 and 𝐣 are unit vectors, and
they act perpendicular to one another. And we generally say that the unit
vector 𝐢 acts in the positive 𝑥-direction, whilst the unit vector 𝐣 acts in a
positive 𝑦-direction.
We’re actually being asked to
calculate some information about the resultant of these forces. And we recall of course that the
resultant of a number of forces is simply the vector sum of those forces. So we’re going to add together the
vector 𝐅 sub one, 𝐅 sub two, and 𝐅 sub three. We can therefore say that the
resultant of our three forces is going to be a single vector in terms of 𝐢 and
𝐣. We calculate this vector by adding
negative four 𝐢 plus two 𝐣, five 𝐢 minus seven 𝐣, and two 𝐢 plus nine 𝐣.
Now, of course, to add a number of
vectors, we simply add their individual components. So we’ll begin by looking at the
𝐢-components. That’s negative four 𝐢 plus five
𝐢 plus two 𝐢. Negative four plus five plus two is
three. So we know the 𝐢-component of our
resultant is three. Then we add the 𝐣-components,
that’s two, negative seven, and nine, giving us a 𝐣-component of four for our
resultant vector. And of course this is in newtons
still. And so that’s the resultant of our
forces.
But in fact, we were looking to
find 𝑅, the magnitude of the resultant. And we might recall that, as a
direct result of the Pythagorean theorem, the magnitude of a vector whose
𝐢-component is 𝑎 and whose 𝐣-component is 𝑏 is the square root of the sum of the
squares of the individual components. So it’s the square root of 𝑎
squared plus 𝑏 squared. And this, therefore, means that the
magnitude of our resultant will be the square root of three squared plus four
squared. Well, three squared plus four
squared is 25. So we’re finding the positive
square root of 25, which is simply five. The resultant vector is in newtons,
and so the magnitude of this vector must also be in newtons. And we can, therefore, say that the
magnitude of the resultant is five newtons.
We still need to find the value of
tan of 𝜃, where 𝜃 is the angle that our resultant makes with the positive
𝑥-axis. Let’s sketch this out and just see
what it might look like. Here is our resultant vector. It’s three 𝐢 plus four 𝐣. Since the horizontal component, the
𝐢-component, is three units and the vertical component 𝐣 is four units, we can add
a right-angled triangle with the measurements three and four units as shown. Of course, we’ve worked out the
resultant, and that’s equal to five newtons. 𝜃 is this angle here. It’s the angle that our resultant,
the yellow line, makes with the positive 𝑥-axis.
We have a right-angled triangle for
which we actually know three of its sides. And so to find the value of the
included angle, we’re going to use right angle trigonometry. We then see that the side adjacent
to the included angle is three units and the side opposite to it is four units. We can therefore use the tan ratio
to work out the value of 𝜃. Since tan of 𝜃 is opposite over
adjacent, here tan of 𝜃 must be equal to four-thirds. And in fact, we’re finished.
We could, if we were required to,
work out the value of 𝜃 by finding the inverse or arc tan of both sides of our
equation. But we were actually asked to find
the value of tan of 𝜃, and so we’re done. The value of 𝑅 is five newtons,
and tan of 𝜃 is equal to four-thirds.
