### Video Transcript

The figure shows a uniform wire π΄π·. It has been bent at π΅ and πΆ to form right angles. The wire was freely suspended from π΄. Find the measure of the inclination angle π΄π΅ to the vertical when the body is hanging in its equilibrium position. Round your answer to the nearest minute.

Recall that when we suspend a rigid body from a point in its equilibrium position, its center of mass will be directly below that point. In this example, we are suspending the body from the point π΄. We can roughly imagine that the body will rotate clockwise about π΄ before settling to its equilibrium position. And as a rough guess by I, the body center of mass should be about here. When the body reaches its equilibrium position, its center of mass will be directly below π΄. So, this dotted line will be the vertical. And we are asked to find the angle between π΄π΅ and the vertical, this angle here π.

We can find this angle first by finding the horizontal distance of the center of mass from π΄, π₯ and then the vertical distance of the center of mass from π΄, π¦. This angle here is also equal to π, since it is enclosed by the same line between π΄ and the center of mass and another vertical line. We can find π using trigonometry with the values of π₯ and π¦. So, first we need to find the horizontal and vertical distance of the center of mass from the point π΄. To do this, we first need to find the masses and the center of masses of the three rods that make up π΄π·.

Since π΄π· is uniform, it does not actually matter what the density of the wire is. So, letβs assume that we have a density of one kilogram per centimeter of wire. The mass of π΄π΅ is therefore 49 kilograms, the mass of π΅πΆ is 36 kilograms, and the mass of πΆπ· is 21 kilograms. To find the center of mass of the whole wire, we can model each of these three rods as a particle. Recall that the center of mass of a uniform rod is at the midpoint of the rod.

Going back to our diagram, letβs define a coordinate system with positive π₯ to the right and positive π¦ vertically downwards with π΄ at the origin of the coordinate system. So, for example, the π¦-coordinate of the point π΅ is 49. The center of mass of the rod π΄π΅ is at the midpoint of π΄π΅, which is vertically below π΄. So, it has an π₯-coordinate of zero and a π¦-coordinate of 49 divided by two. Similarly, the center of mass of π΅πΆ has a π¦-coordinate of 49 and an π₯-coordinate of 36 divided by two, which is equal to 18. And finally, the center of mass of πΆπ· has an π₯-coordinate of 36 and a π¦-coordinate of 49 minus 21 over two. And this simplifies to 77 over two.

We now have everything we need to find the center of mass of the whole wire. Letβs clear a little space first. And recall that the π₯-coordinate of the center of mass of a system of particles COM π₯ is given by the sum of all the products of the particlesβ masses, ππ, times their π₯-coordinate, π₯π, all divided by the total mass of all the particles. Substituting in the masses of the rods and the π₯-coordinates for their centers of mass, we get 49 times zero plus 36 times 18 plus 21 times 36 all divided by the total mass 49 plus 36 plus 21. This all comes to 1404 divided by 106, which is the π₯-coordinate at the center of mass of the wire.

Now, we need to do the same thing for the π¦-coordinates of the rod. So again, substituting in the masses of the uniform rod and their π¦-coordinates gives us 49 times 49 over two plus 36 times 49 plus 21 times 77 over two all divided by 49 plus 36 plus 21. This all comes to 3773 divided by 106, which is the π¦-coordinate of the center of mass of the wire.

Now, to find the angle π that the line π΄π΅ makes with the vertical, we use tan π equals the opposite π₯ over the adjacent π¦. π is therefore given by the inverse tangent of the center of massβs π₯-coordinate divided by its π¦-coordinate. The identical denominators will cancel, giving us the inverse tan of 1404 over 3773. This gives us our final answer. π equals 20.41 degrees, which is equal to 20 degrees and 25 minutes.