1. The problem statement, all variables and given/known data a man holds a 5.00kg weight in his hand. His arm is represented here by a pair of thin rods attached with a hinge, with a cord for the bicep muscle. The forearm makes a 20 degree angle with the horizontal, and the bicep (which is attached 4.0 cm away from the elbow joint) makes a 55 degree angle from the horizontal. assumme that the forearm has a mass of 1.2 Kg and is 36cm long, and has its centre of mass midway through its length. find the tension in the bicep muscle and the components of force in the elbow joint! DIAGRAM: http://farm9.staticflickr.com/8445/7797836656_015d170c93.jpg 2. Relevant equations Torque =F*d 3. The attempt at a solution Okay so ill show you two ways i've don't it because I'm confused about how its actually done 1st way it think it may be wrong Sin(55)(4)Ft=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36) solveing for Ft = 556N Attempt two sin(55+20)*4(ft)=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36) Ft= 480N and the vertical component = 9.8*1.2+5*9.8-480*sin55=-332N or 332N to the vertical horizontal component = sin55*480=393N Am i going wrong anywhere, which method is correct PLEASE HELP!