The mass of the arm is 2270g.(Angle:90) The mass of the arm is 2890g.( Angle:50) The mass of the arm is 2930g. ( Angle:130) The mass of the arm is 4325g.( Angle:50) M, = 150g The mass of the arm is 3595g.( Angle:70) My = 150gShow your work for calculating the force at the elbow joint (F) for 0 = 50. and My = 0 g. Do not include an uncertainty.\fThese torques must sum to zero so 213,- = le sing ng(d1+ (12 + d3) - mA8(d1+ d2) = 0 (\"-3) J If we use the pulley system as outlined in Figure 11.1, we know that T : ng. We can then substitute this into Equation 11.3 and cancel out a g in each term. delsinSmL(d1+d2+d3)mA(d1+d2) =0. (11.4) This equation governs the statics of the arm. For the experimental setup, we adopt - d1 : 4.1 :l: 0.1 cm as the distance from the elbow joint to where the bicep connects to the bones in the forearm (radius and ulna) through the distal biceps tendon (here represented by a string). - (11 + d; = 16.0 :t 0.1 cm is the distance from the elbow to the centre of the arm where the weight of the arm acts on the system. - d1 + d2 + d3 = 31.4 :I: 0.1 cm is the distance from the elbow to the where the load acts. Force Forces and Torques T CF. = 0 Load MT Elbow Joint ML FEx = Ta Figure 11.1: Physics Model of the Human Arm CF = 0 TT FE,x X - - - - FE,y + FL + FA = Ty - - - - - E , y d2 d3 FE m'Ag mig IFE = FEy + FE,x Figure 11.2: Force human