Forces in 2D Legacy Problem #16 Guided Solution

An effective problem solver reads the problem carefully and develops a mental picture of what's going on, identifies the known and the unknown values, and then plots out a strategy as to how to get from the known to the unknown quantity. Here we have a picture of a chandelier that's hanging from four cables. The mass of the chandelier is 2,900 kilograms. Just to give you the feel for the massiness of the chandelier, that would be equivalent to a little bit more than 6,200 pounds. Now what we know is that all the forces balance each other. That is, the downforce of gravity acting up on the chandelier is balanced by the four individual upforces supplied by each cable that supports the downforce. That is, that 4 times the single FY force is going to be equal to the F graph force. Now I can calculate the F graph simply by going M times G, where the M is 2,900 kilograms and the G is 9.8 newtons per kilogram. Since I know the downforce, the force of gravity, I can divide that value by 4, and what I would then have is the force in one of the cables, the vertical force in one of the cables. Now what I will need to do is, focusing on one cable, draw the force triangle with a tension force going at a 63 degree angle with respect to the horizontal. It has two components, one horizontal and one vertical, and I would just sketch a little right triangle there showing the horizontal and the vertical components of this tension force. Now I want to relate the tension force to the vertical pole of the tension force, which I have calculated from the weight of the sign. The vertical pole would be 7,105 newtons, and that would be equal to the sign opposite the 63 degree angle. So I say the sign of 63 degrees is equal to 7105 newtons divided by F tension, and I solve for F tension. I will have to do that carefully using good algebra skills, multiplying both sides of the equation by F tension and then dividing through by the sign of 63 degrees. That would give me a value of 7,974 newtons, which is way too many digits, so rounding to two significant figures, I would have 8.0 times 10 to the third newtons.