Video Transcript
Coplanar forces of magnitudes 𝐹
newtons, eight root three newtons, root three newtons, and nine root three newtons
act on a particle, as shown in the diagram. Given that the magnitude of their
resultant is nine root three newtons, determine the value of 𝐹.
We’re given information about the
magnitude of the resultant of four coplanar forces, where of course the resultant is
the vector sum of those forces. The problem is, one of those forces
is 𝐹, and that’s what we’re trying to find. So what we’re going to do is
resolve each of our forces into perpendicular components. And then we’re going to find the
vector sum of these and then find the magnitude and equate that to nine root
three. So let’s define the direction in
which the 𝐹 force is acting to be the positive horizontal direction. And then perpendicular to this and
upwards is the positive vertical direction.
Let’s begin by resolving our forces
then in a horizontal direction. We know 𝐹 is acting in this
direction. But actually, we need to find the
horizontal component of the eight root three, nine root three, and root three
forces. We add a right-angled triangle to
our eight root three force. We know that the horizontal
component will be acting to the right and the vertical component this will be acting
upwards. And then we can use right angle
trigonometry to see that the adjacent side, and that’s the horizontal component, is
eight root three cos 60 and the opposite side, the vertical component, is eight root
three sin 60.
Let’s repeat this for the root
three force. Since angles on a straight line add
up to 180 degrees, the included angle this time is 60. And then we see that the side
opposite to this, that’s the vertical component acting upwards, is root three sin
60. And the adjacent, which is the
horizontal component acting to the left, is root three cos 60. There’s one more triangle that
we’re interested in. Finally, we have the nine root
three force. We could add a triangle to the left
of this, but there’s an awful lot going on there already. So we add a triangle to the right
as shown.
We know that this angle here is 60
degrees. So the included angle must be
30. And once again, we resolve it into
its horizontal and vertical components. We get nine root three cos of 30
for the vertical component and nine root three sin of 30 for the horizontal
component.
Now that we’ve resolved all of
these forces, let’s find the horizontal sum. We have 𝐹 acting in a positive
direction plus eight root three cos of 60. That’s the component of the eight
root three force that acts in the horizontal direction. And then we subtract root three cos
of 60 since that’s acting to the left, and nine root three sin of 30. By using the fact that both sin of
30 and cos of 60 are one-half, we simplify this to 𝐹 plus eight root three times
one-half minus root three times one-half minus nine root three times one-half. And all that rather complicated
working simplifies to just 𝐹 minus the square root of three.
Let’s do the same in the vertical
direction. We know that we have eight root
three sin 60 acting upwards, so in the positive direction, and root three sin
60. But in the opposite direction, we
have nine root three cos of 30. So we’re going to subtract this
from our sum. This time, we use the fact that sin
of 60 and cos of 30 are root three over two. And so we get eight root three
times root three over two plus root three times root three over two minus nine root
three times root three over two, which is in fact equal to zero. And that’s really useful because we
can actually work out the magnitude of the resultant really easily.
Since our force is only acting in
one direction, the magnitude of our resultant must be 𝐹 minus root three. But we were told that the magnitude
of the resultant is nine root three. So we can set up a simple equation
for 𝐹. 𝐹 minus the square root of three
is equal to nine root three. And we’ll solve for 𝐹 by adding
root three to both sides. And so we see that 𝐹 is 10 root
three or 10 root three newtons.
