A solid uniform 45.0-kg ball of diameter 32.0 cm is supported against a frictionless wall using a 30.0 cm wire of negligible mass. (a) find

"2. Relevant equations ∑ ( F x ) = 0 ∑(Fx)=0 ∑ ( F y ) = 0 ∑(Fy)=0 a 2 + b 2 = c 2

Now, using the equilibrium equations, I got ∑ ( F y ) = − 441.45 + T S i n θ = 0 ∑(Fy)=−441.45+TSinθ=0 So, T s i n θ = 441.45 Tsinθ=441.45 (Newtons). Now for the x components. ∑ ( F x ) = T c o s θ − N = 0 ∑(Fx)=Tcosθ−N=0 So, T c o s θ = N Tcosθ=N Now I need to figure out what the angle is. The diameter of the sphere is 32cm, so the radius must be 16cm. The length of the wire is 30cm. I used the pythagorean theorem to find the height from the sphere to the wire. 16 2 + b 2 = 30 2 162+b2=302 256 + b 2 = 900 256+b2=900 b 2 = 644 b2=644 b = 25.38 b=25.38 Now arcsin ( 25.38 / 30 ) = 57.78 d e g r e e s arcsin⁡(25.38/30)=57.78degrees This is where I start to have trouble. I figured I would find T by plugging that angle into T s i n θ = 441.45 Tsinθ=441.45 and solving for T. This gave me an answer of 521.8 N. When I check the back of the book, the answer is actually 470"