Question

3. The figure shows an outstretched arm with a mass of 4.2 kg. The arm is 56 cm long, and its center of gravity is 21 cm from the shoulder. The hand at the end of the arm hold a mass of 6.0 kg. (a) What is the torque about the shoulder due to the weights of the arm and the 6.0 kg mass? (b) If the arm is held in position by the deltoid muscle, whose force on the arm acts 5.0 below the horizontal at a point 18 cm from the shoulder joint, what is the force exerted by the muscle? (Ans. 40.2 Nm, 1.28 kN) 56 cm 21 cm 6.0 kg 15°

Answer 1

draw the free body diagram for the given let us fost sctuation. freebody diagram: 18cm 56 cm 24cm Ploso 5o 150 3 Dioso In} To mg end of the arm Mg = weight ong= weight of the man at the of the arm. Can Torque about the shoulder: first of all let y ford the tongue about the shoulder, due to the weight of the arm .: 2 - (AC) (ng sca 105) [ corque = Forcek distance) - (21cm) (4.2kg) ca.8/52) son 1050 - (0.21n) (4.2kg) (9.8 / 2) sin roso - 8.35 NM Now, tongue about the shoulder due to 6 log onase at D. 2 : (AD) ( 9 sen Logº) - (0.56 m)( 68 9.8k Son 1050 ) - 32.81 Nm Net torque is: Inet = Elt tz a cat 8-35 t 31-81 10.16 – 40.2 NM (6) Now, the arm is held by the deltoid muscle .. let, the force emented by the murdle is F then the torque produced by this force about the shoulder = (AB) CF sin 170°) = (0.18 m) (F son 170°) [see the For static equillibrum, This torque is equal to ener .: (0.18) Esco 170° - 40.2 NM = 46.2 = 12 86 .12 Newton free body diagram] Co. 8) (sch 1700) = 1.286 KN.