Question

please answer in full detail

4. Use the drawing below as a model of forces involved in holding an outstretched arm stationary, let Fu model the "deltoid muscle, which provides tension at an angle of to the bones and _cm represent the center of mass of the arm. X_M12 cm x cm 24 cm mass of the arm - 3.3 kg 015 a. the shoulder joint labeled in the above figure but decomposed into it's - and y. components in the free body diagram below) is the pivot, draw the direction of torque for each forces Fm and Fcm in the figure below. (2 points) FLY FM AT b. Write an equation for the net torque on the outstretched arm and solve for Fu. (8 points) ST= --4/2 + FMLSINA X - 12/2+ Firsin5 :0 122 Far For 5.04 = 1 c. Write an equation for the net Force in the x-direction and in the y direction in terms of Fuand the x-and y. components of the force on the pivot (F). What is the magnitude of force on the shoulder joint. F? (10 points) F = FastF-Y-O E F y = Ffy na Mesins=0 + 2 v SnoL d. Suppose the hand now holds a 16 kg mass. What force, Fm, is required of the deltold muscle assuming the mass is 52 cm from the shoulder joint? (5 points) F-4 - Me, PwC - (1 6kg) (9.8) + FM (252X5115)= 156.8. 152[Sinis) FM= 7860901

Answer 1

For Emsing cm b) Here net torque will be zero fine for equilibrium position hence Etze here Ił = En sind) x Xon - (F.m.) X Xcm - or (CFM) sinox = Fem Xem - For Hem - 33x9.8X24 2250 N or From = tem sino = 12x siniso ov 12 x siniso or Fn =250N c) Here for equilibrium E Fx =0 and E Fy=0 Z Fx = FR - FM COSO 20 or Fix = F COSO or Fix = 250 Cos 15° = 242 N and I fy = Fig + Frn sino - Fcm = 0 or a Fem - Ensino = 3.3x9.8+-250 sinis Ejy = or Fiy = -32.3 N d) Now for equilibrium = 0 ST = F (sino) xm - Fem XXcm. - (16x9-81 ) x 52 OYF-24/3.3x931)+(1679.8152 - 2878 N 12 x sinis = 48 TO N