Video Transcript
𝐴𝐵𝐶 is a triangle, where the
side 𝐴𝐵 equals 20 centimeters, 𝐵𝐶 equals 25 centimeters, and 𝐶𝐴 equals 15
centimeters. And forces of magnitudes 120, 150,
and 90 newtons are acting along 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴, respectively, where the system
is equivalent to a couple. Determine the magnitudes of the two
parallel forces that would make the system in equilibrium when acting at 𝐵 and 𝐶
perpendicularly to 𝐵𝐶.
Let’s start with a diagram of the
scenario. We have the triangle 𝐴𝐵𝐶 with
side lengths 𝐴𝐵 equals 20 centimeters, 𝐵𝐶 equals 25 centimeters, and 𝐶𝐴 equals
15 centimeters. The force along 𝐴𝐵 is 120
newtons. The force along 𝐵𝐶 is 150
newtons. And the force along 𝐶𝐴 is 90
newtons. We have two parallel forces 𝐹 one
and 𝐹 two, acting from the points 𝐵 and 𝐶 and perpendicular to 𝐵𝐶. The other forces acting on the
triangle are equivalent to a couple. So in order to bring the system to
equilibrium, the two forces acting at 𝐵 and 𝐶 must also be equivalent to a couple
of equal magnitude and opposite direction. So 𝐹 one is equal to 𝐹 two. Let’s call this magnitude just
𝐹.
To find the magnitude 𝐹, we need
to find the magnitude of the couple formed by the other forces acting on the
triangle. To do this, we can calculate the
moments of all three forces about a point of our choice and then add them
together. Let’s calculate the moments of the
three forces about the point 𝐴, taking the convention that counterclockwise moments
are positive.
Recall that the magnitude of the
moment 𝑀 of a force 𝐹 acting from a point 𝑝 is equal to the magnitude of the
force 𝐹 multiplied by the perpendicular distance 𝑑 between the pivot point 𝑜 and
the line of action of 𝐹. The lines of action of the forces
of 90 newtons and 120 newtons pass through the point 𝐴. So their perpendicular distances
from 𝐴 are both zero. And hence, their moments about 𝐴
are also both zero. The only force that has a nonzero
moment about 𝐴 is the force of 150 newtons. We need to find the perpendicular
distance from the line of action of this force to the point 𝐴. This is equivalent to finding the
altitude of the triangle with the line 𝐵𝐶 as the base.
Recall that the altitude of a
scalene triangle, that is, a triangle with three sides of different lengths, is
given by two over 𝑏 multiplied by the square root of 𝑠 times 𝑠 minus 𝑎 times 𝑠
minus 𝑏 times 𝑠 minus 𝑐, where 𝑠 is the semiperimeter of the triangle. 𝑏 is the base of the triangle. And 𝑎 and 𝑐 are the other two
sides of the triangle. The semiperimeter, 𝑠, is half the
perimeter of the triangle, which is equal to 15 plus 20 plus 25 over two, which is
equal to 30 centimeters. 𝑏 is the length of the base line
𝐵𝐶, which is 25 centimeters. 𝑎 and 𝑐 are the lengths of the
other two sides, which are 20 centimeters and 15 centimeters.
So, the altitude of the triangle,
which is equivalent to the perpendicular distance 𝑑 between the pivot point 𝐴 and
the line of action of the force of 150 newtons, is given by two over 25 multiplied
by the square root of 30 times 30 minus 20 times 30 minus 25 times 30 minus 15. This comes to the nice round
number, 12 centimeters. The magnitude of the moment 𝑀 is
therefore given by the magnitude of the force, 150, multiplied by this perpendicular
distance, 12, which is equal to 1800 newton-centimeters. Therefore, the magnitude of the
resultant moment on the system is 1800 newton-centimeters. And the magnitude of the couple of
the two parallel forces introduced at the points 𝐴 and 𝐵 must therefore also be
1800 newton-centimeters.
Recall that the magnitude of a
couple acting at two points 𝐵 and 𝐶 with the forces at an angle 𝜃 between their
lines of action and the line 𝐵𝐶 is equal to the magnitude of one of the forces,
𝐹, multiplied by the length of the line 𝐵𝐶 multiplied by its sin 𝜃. In this case, the two forces are
perpendicular to the line 𝐵𝐶. So the angle 𝜃 is equal to 90
degrees, and therefore sin 𝜃 is equal to one. We can therefore simplify this to
𝐹 times the length of the line 𝐵𝐶. We can rearrange this equation for
𝐹 to give 𝐹 equals the magnitude of the couple, 𝑀 𝐵𝐶, over the length of the
line 𝐵𝐶. This is equal to 1800 over 25. Therefore, the magnitudes of the
two parallel forces that would make the system in equilibrium are both 72
newtons.
