Video Transcript
A thin uniform wire is shaped
into trapezoid 𝐴𝐵𝐶𝐷, where the measure of angle 𝐵 equals the measure of
angle 𝐶, which equals 90 degrees; 𝐴𝐵 equals 494 centimeters; 𝐵𝐶 equals 105
centimeters; and 𝐶𝐷 equals 134 centimeters. Find the distance between the
center of gravity of the wire and point 𝐵, rounding your answer to two decimal
places.
Before we attempt to answer
this question, we’re going to begin by sketching out trapezoid 𝐴𝐵𝐶𝐷. We know that the measure of
angle 𝐵 and the measure of angle 𝐶 are 90 degrees. This tells us that side lengths
𝐴𝐵 and 𝐶𝐷 must be the parallel sides of the trapezoid. Let’s add the trapezoid to the
coordinate plane. And we’ll let point 𝐵 be at
the origin, since that’s the point we’re going to be working with a little later
on.
Now, before we go any further,
we’re actually going to calculate the length of line segment 𝐴𝐷. And we’ll see why that’s
important in a minute. By adding a right triangle to
the diagram, we can in fact use the Pythagorean theorem to calculate the value
of 𝑥. This tells us that the square
of the hypotenuse, so 𝑥 squared, is equal to the sum of the squares of the two
shorter sides. Well, the base of the triangle
is in fact 105 centimeters. And its height is the
difference between 494 and 134. So 𝑥 squared is 140625. The square root of this gives
us a length of 375 centimeters.
Now this is really useful
because we’re told that the wire is uniform, so each rod has the same
density. And we can define their linear
density to be one kilogram per centimeter. And we can therefore say that
since the rods have a constant mass distribution, their masses are proportional
to their length. And so the mass of the piece of
wire between 𝐴 and 𝐵 is 494 kilograms, 𝐵𝐶 is 105 kilograms, 𝐶𝐷 134
kilograms, and 𝐴𝐷 375 kilograms.
The next thing that we’re going
to do is work out the center of mass of each rod. We know that the center of mass
is located at the midpoint of each rod. So we’ll work out the
coordinates of each midpoint. Line segment 𝐴𝐵 lies on the
𝑦-axis, so the 𝑥-coordinate of its midpoint will be zero. Then its 𝑦-coordinate is the
average of the 𝑦-coordinate of 𝐴 and 𝐵. So that’s 494 plus zero divided
by two, which is 247. Then, line segment 𝐵𝐶 lies on
the 𝑥-axis. So the 𝑦-coordinates of its
midpoint is zero. Then, the 𝑥-coordinate of its
midpoint is half of 105, say 52.5.
To get to the midpoint of 𝐶𝐷,
we move 105 units along the 𝑥-axis. And then we move a half of 134
units up. So its 𝑥-coordinate is 105 and
its 𝑦-coordinate is 67. Then, we have the midpoint of
𝐴𝐷. The midpoint of its
𝑥-coordinate is, in fact, 52.5. Once again, it’s a half of
105. Then the 𝑦-coordinate of its
midpoint is, in fact, the average of 494 and 134, which is 314. Now, we recall that the
coordinates of the center of mass are the average of each particle’s coordinates
weighted with its mass.
So we’ll begin by calculating
the 𝑥-coordinate of the center of mass of the entire system. Then we find the numerator by
finding the sum of the products of each element in the first and second row of
our table. The denominator is just the
total mass of the system, so it’s the sum of all elements in the first row
itself. That gives us 39270 over
1108.
Let’s clear some space and
repeat this for the 𝑦-coordinate of the center of mass. To find the numerator of the
𝑦-coordinate of the center of mass, we find the sum of the products of the
elements in the first row and the third row. And then, once again, we divide
that by the total mass. That gives us 248746. And so we have the coordinates
of the center of mass. Of course, we want to find the
distance of this point from point 𝐵. And we know that point 𝐵 is
the origin. And so we’re going to use the
distance formula, which is, of course, just a special version of the Pythagorean
theorem.
The distance is simply the
square root of the sum of the squares of the 𝑥- and 𝑦-coordinates, which is
227.28. The distance between the center
of gravity of the wire and point 𝐵, correct to two decimal places, is 227.28
centimeters.
