Hello, I wanted to make sure that I am solving this problem in the correct manner, because one of my answers seems a little off from the book's. "Consider a conical pendulum with an 80.0-kg bob on a 10.0-m wire making an angle of 5.00(degrees) with the veritcal. Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration on the bob." Here are my steps in solving this problem: I drew a free-body diagram of the bob and the string and noticed immediately that above the bob (the string) is tension, and that tension is causing the centripetal force, and the bottom of the bob is mg (weight). I realized that a right triangle could be drawn in this scenario. The hypotenuse is equal to 10 m, because that's the length of the wire. I broke down the scenario into x (or r) and y force components. Fx(or Fr) = T(sin5) = m(v^2/r) Fy = T(cos5) - mg = 0 From here, I went ahead and solved for T. So T(cos5) = mg, which equals 787 N. After that, I plugged the T value into the Fx(or Fr) equation and got 68.6 N. The book agrees with me that the x component of the force is 68.6 N, but it says for 784 N for the y component. My answer is three Newtons off. That seems too much? In solving for (b), I divided the two force equations together, which cancelled out m and converted the sin and cos into a single tan. I knew I would need to know the values for velocity, and the radius to continue. First, I solved for the radius using my right triangle and basic geometry to realize that the radius = the hypotenuse multiplied by sin5. Because the hypotenuse = 10 m, my radius = 0.872 m. After that, I solved for velocity, which = 0.864 m/s. To find the radial acceleration, I divided the two (with velocity being squared) and got 0.856 m/s^2. The book agrees with me there as well. To check my radial acceleration answer, I made sure that it was equal to Tsin5, and it is. Is that the correct way in solving this kind of problem? Any opinions or suggestions would be great. Also, is it acceptable to be off by three Newtons in the answer above? Thank you.