Forces in 2D Legacy Problem #15 Guided Solution

An effective problem solver will read a problem carefully and get a mental picture of what's going on. We'll record what is known and identify the unknown value applied on a strategy as to how to get from the known to the unknown quantities. Here we have a flower pot that's being hung from the ceiling by three chains. The picture that I have is of a flower pot and three forces pulling up on the flower pot. I'm told that the chains make an angle of twenty-five degrees, and the next three words are important, twenty-five degrees with the vertical. In other words, twenty-five degrees with the vertical, and sixty-five degrees with the horizontal. I'm to determine the tension in one of the chains, and since all chains are oriented at the same twenty-five degree with the vertical angle, all three chains pull with the same tension. So the concepts that I'm going to use to get myself from the known quantities to the unknown quantities is the idea that this flower pot is hanging at equilibrium, and as such, all of the forces will balance. There are three up forces and one down forces. The three up forces are simply the vertical components of the tension force in each chain. And the down force is the force of gravity acting upon this 5.6 kilogram mass. Now I can find the down force by going m times g, and once I do, I know that each chain pulls upwards with one-third of this down force. That would make all the forces balance. So once I know the vertical pull in each of the chains, or I should say in just one of the chains, I can use the twenty-five or the sixty-five degree angle and determine the diagonal pull, or the force of tension. So here I go. I begin with the 5.6 kilograms, which is the mass of the pot, and the down force would be m times g, which comes out to be 54.88 newtons. Now this 54.88 newtons is the total up force in those three chains, so I'm going to have to divide by three. I get 18.2933 newtons as the up force in each one of the chains. Now that's up, not up and diagonal, but up, and I want to find the diagonal force. So I draw myself a force triangle. I draw a tension force in an angle of sixty-five degrees with the horizontal, twenty-five with the vertical, and then I make a triangle, a right triangle out of that tension force. It's the hypotenuse of a right triangle that has as its horizontal side an Fx, and as its vertical side a Fy of 18.2933. Now you know that this Fy is the side opposite a sixty-five degree angle in a force triangle, or if you do it another way, it's the side adjacent the twenty-five degree angle. Considering it to be the side opposite the sixty-five degree angle, I could say that the sine of sixty-five is 18.2933 over the tension. I need to solve this equation for tension, so I multiply each side by F tension and divide through by the sine of sixty-five, and when I'm done I get 20.1845 newtons. I can round that to the proper number of significant digits. It's twenty exactly newtons. Now, doing it alternatively with a different force triangle, maybe you drew the tension force in a twenty-five degree angle between the Fy, or vertical side, and the tension force, and if you did, you'd do it a little bit differently, and simply say the cosine of twenty-five degrees is the ratio of the side adjacent, which would be Fy, divided by F tension. Cosine twenty-five equals 18.2933 all over F tension. Multiplying for F tension gives you the same 20.1845 newtons, and it matters not how you draw that triangle, as long as you use the appropriate trig function.