Please let me know if I did this wrong or right, and if I did it wrong, please correct me :) 1. The problem statement, all variables and given/known data The biceps brachii, a muscle in the arm, connects the radius, a bone in the forearm, to the scapula in the shoulder (see below). The muscle attaches at two places on the scapula but at only one on the radius. To move or hold the arm in place, the biceps muscle balances the weight of the arm and the force at the elbow joint. The centre of mass of the arm is at 15 cm from the elbow joint. The horizontal force of the elbow joint is 6.5 N when the forearm is held parallel to the ground and the forearm weighs 15.3 N. If the biceps supports the entire weight of the forearm, calculate the necessary force from each branch of the biceps to hold the forearm parallel to the ground and the vertical force at the elbow. 2. Relevant equations net F = 0; net Fx = 0, net Fy = 0 net torque = 0 3. The attempt at a solution Choose E as point of rotation (i.e., Moment Arm of E, MAE = 0) MAarm = (15.3 N)(15 cm) = 229.5 N cm Determine angles of insertions of A and B: Firstly, I assume that, based on the diagram, points E and B (the point where B attaches to the scapula) line up vertically - is this assumption valid? Thus angle of insertion of A = thetaA = tan-1(30/2) = 86.19 deg Similarly, thetaB = tan-1(30/5) = 80.54 deg Torque balance: MAarm = MAA + MAB 229.5 N cm = FA * (5 cm)sin(86.19 deg) + FB * (5 cm)sin(80.54 deg) Rearranging for FA, FA = [229.5 N cm - FB * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) ... Eq. 1 Note: The "(5 cm)sin(86.19 deg)" and "(5 cm)sin(80.54 deg)" represent, respectively, the lengths of the lines that originate at E and intersect the FA and FB vectors such that the lines form right angles with said vectors. Force balance: x-direction: FEx = FAx + FBx 6.5 N = FAcos(86.19 deg) + FBcos(80.54 deg) FA = [6.5 N - FBcos(80.54 deg)] / cos(86.19 deg) ... Eq. 2 Substituting Eq. 1 into Eq. 2 yields: [229.5 N cm - FB * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) = [6.5 N - FBcos(80.54 deg)] / cos(86.19 deg) 46 N - 0.989FB = 97.82 N - 2.473FB 1.484FB = 51.82 N FB = 34.92 N Substituting this result into either of Eqs. 1 or 2 yields FA = 11.47 N y-dir'n force balance: 15.3 N = FAy + FBy + FEy FEy = 15.3 N - (11.47 N)sin(86.19 deg) - (34.92 N)sin(80.54 deg) FEy = -30.6 N - am I supposed to get a negative answer here?!