Video Transcript
A uniform lamina is in the form of
a rectangle 𝐴𝐵𝐶𝐷, where 𝐴𝐵 equals 56 centimeters and 𝐵𝐶 equals 35
centimeters. Two points 𝐸 and 𝐹 are on 𝐴𝐵
such that 𝐴𝐸 equals 𝐵𝐹 equals 14 centimeters. The triangle 𝑀𝐸𝐹, where 𝑀 is
the center of the rectangle, is cut out of the lamina. Find the coordinates of the center
of mass of the resulting lamina. Given that the lamina was freely
suspended from 𝐷, find the tangent of the angle that 𝐷𝐴 makes to the vertical,
tan 𝜃, when the lamina is hanging in its equilibrium position.
In this question, our first task is
to find the center of mass. In this scenario, we effectively
have two laminas. We have the rectangle with its
center of mass at its geometric center 𝑀 and we have this missing triangle with its
own center of mass at its geometric center. The rectangular lamina has a mass
of 𝑚 one which is the mass of the lamina before we remove the triangle. The missing triangle has a mass of
𝑚 two. Using the negative mass method, we
can treat this triangle as a separate lamina with a negative mass of negative 𝑚
two. We can model these two separate
lamina as two particles of mass 𝑚 one and negative 𝑚 two located at the centers of
mass of the laminas.
Recall that for the center of mass
of two particles, the 𝑥-coordinate is given by the products of the masses and their
𝑥-coordinates divided by the total mass and likewise for the 𝑦-coordinates. In this case, the 𝑥-coordinate of
the center of mass is trivial. Looking at the description, 𝐴𝐸 is
equal to 𝐵𝐹 which is equal to 14 centimeters. Therefore, the triangle has a base
length of 56 minus 14 minus 14, which is equal to 28 centimeters. The base of this triangle is
therefore centered at the midpoint of the top side of the rectangle. Since the opposite corner touches
the rectangle’s midpoint, this is therefore an isosceles triangle. Its center of mass is therefore
directly above the midpoint of the rectangle.
We therefore have a line of
symmetry in the lamina, and the 𝑥-coordinate of both centers of mass is halfway
along the side of the rectangle, so 56 over two which is equal to 28. For the 𝑦-coordinate, we will need
to do the complete calculation. Firstly, the particle modeling the
missing triangular lamina has negative mass. We therefore need to replace these
plus signs in the formula with minus signs. Next, we need to find the masses of
the two particles. Since the lamina is uniform, its
density is constant throughout. Therefore, the mass of the
rectangular lamina and the triangular lamina will be directly proportional to their
areas.
In the question, we are told that
𝐵𝐶 is equal to 35 centimeters. Therefore, the area of the whole
rectangle is 56 times 35 which is equal to 1960 centimeters squared. For the triangle, we know that its
bottom corner touches the midpoint of the rectangle. Therefore, its altitude is 35 over
two centimeters. The area of the triangle,
therefore, is given by one-half times its base times its altitude, which gives us
one-half times 28 times 35 over two, which is equal to 245 centimeters squared.
Now, we need to find the
𝑦-coordinates of the centers of mass of the two separate laminas. For the center of mass of the
rectangle, this is easy. Its 𝑦-coordinate is halfway up the
rectangle, which is 35 over two. Now, recall that the geometric
center of a triangle is one-third of the way along the line connecting the middle of
the base and the opposite corner. The distance between the top of the
rectangle and the triangle center of mass, therefore, is one-third times the
triangle’s altitude, 35 over two, which simplifies to 35 over six. The 𝑦-coordinate of the center of
mass of the triangle, therefore, is 35 minus 35 over six, which comes to 175 over
six.
We now have everything we need to
find the 𝑦-coordinate of the center of mass of the whole lamina. The masses of each lamina are given
by the density, 𝜌, times their areas, 𝐴 one and 𝐴 two. But since the density is uniform,
𝜌 is constant, and we have a common factor of 𝜌 in every single term which we can
cancel. This leaves us with the area of the
rectangular lamina, 1960, multiplied by the 𝑦-coordinate of its center of mass, 35
over two. And since we are treating the
triangular lamina as a negative mass, we subtract the area of the triangle, 245,
multiplied by the 𝑦-coordinate of its center of mass, 175 over six. We then divide by the area of the
resulting lamina, which is 1960 minus 245. This all simplifies to 95 over
six. This gives us our answer the
coordinates of the center of mass of the resulting lamina 28, 95 over six.
Now, we need to find the tangent to
the angle that 𝐷𝐴 makes to the vertical when the lamina is suspended from 𝐷. We just found the center of mass of
the lamina to be 28, 95 over six, which is approximately here. When suspended from 𝐷 and hanging
in its equilibrium position, the center of mass will hang directly below 𝐷. Therefore, this blue dotted line
will be the vertical line. We are asked to find the tangent of
this angle here, the angle between 𝐷𝐴 and the vertical. We can construct a right triangle
by drawing a horizontal line through the center of mass. The tangent of this angle 𝜃 will
be given by the opposite side of the triangle over the adjacent side. These are just the 𝑥- and
𝑦-coordinates of the center of mass. So we have 28 over 95 over six. This simplifies to give us our
final answer tan 𝜃 equals 168 over 95.
