### Video Transcript

A uniform rod of length 50
centimeters and weight 143 newtons is freely suspended at its ends from the ceiling
by means of two perpendicular strings attached to the same point on the ceiling. Given that the length of one of the
strings is 30 centimeters, determine the tension in each string.

Weβre going to begin by drawing a
free-body diagram of this scenario. Letβs define the ends of our rod to
be π΄ and π΅. And so we have our rod suspended
from two perpendicular strings, one of which measures 30 centimeters. Now, in fact, we can work out the
measurement of the second piece of string. Itβs 40 centimeters. And we can calculate that using the
Pythagorean theorem or just recognizing that we have a multiple of a Pythagorean
triple β three, four, five. Since the rod is uniform, we can
say that the downwards force of its weight acts at a point exactly halfway along the
rod. Now, since the rod exerts a
downwards force on the pieces of string, there will be an opposite reaction force of
tension. Letβs call that π sub π΄ for the
tensional force in the first bit of string and π sub π΅ for the tensional force in
the second.

We have our free-body diagram, but
thereβs still an awful lot going on here. So how do we simplify it
further? Well, we know that when three
coplanar forces acting at a point are in equilibrium, they can be represented in
magnitude and direction by the adjacent sides of a triangle taken in order, since
the systems in equilibrium will lay the force vectors end to end to make a
triangle. π sub π΄ and π sub π΅ will be
perpendicular to one another, since the pieces of string along which they run are
also perpendicular to one another. But in fact, with a little bit of
inspection, we can see that these are similar triangles. Weβll define the angle between the
rod and the 30-centimeter piece of string to be π₯ degrees. This is equal to the angle between
the 143-newton force and the tensional force at π΅.

This isnβt hugely intuitive, but we
can convince ourselves that this is true by adding a line parallel to the 143-newton
force on our first diagram and then using the fact that angles in a triangle sum to
180. Since the angles in our force
triangle and the angles in the triangle represented by the lengths of each item are
equal, we know these triangles are similar. And so we can use scale factor or
ratios to find the magnitudes of π sub π΄ and π sub π΅. Letβs consider the ratio of π sub
π΄ to 143. This must be equal to the ratio of
the 40-centimeter length to the 50-centimeter length. And we chose these pairs of sides
in both our diagrams because weβre interested in the pairs of sides that are
adjacent to the angle measuring 90 minus π₯ degrees.

Multiplying both sides of this
equation by 143 and we find that π sub π΄ is 40 over 50 or four-fifths times 143,
which is 114.4. So the magnitude of π sub π΄ is
114.4 newtons. In a similar way, the ratio of π
sub π΅ to 143 newtons will be equal to the ratio of the 30-centimeter length to the
50-centimeter length. To solve for π sub π΅ once again,
we multiply by 143, which gives us 85.8 or 85.8 newtons. The tensional forces in each string
are 85.8 newtons and 114.4 newtons.