Question

Part A - Angular Acceleration of the Rod Learning Goal: To apply the equations of motion to a system that involves rotation about a fixed axis and to use this information to determine key characteristics. The slender rod AB shown has a mass of m = 71.0 kg and is being supported by a rope and pulley system stationed at C. Starting from rest in the position shown), the rope and pulley system tug on the rod causing it to rotate about A. The torque applied to the pulley is T = 2.35 kN.m and has an effective moment arm of r=0.160 m. The dimensions shown in the figure are l = 2.90 m and h = 1.20 m. Assume the pulley is frictionless and massless. Determine the angular acceleration of the rod the instant the rope and pulley system have pulled the rod through an angle of () - 3.10", Express your answer to three significant figures and include the appropriate units. View Available Hint(s) + o ? (Figure 1) rad 88.4 s? Submit Previous Answers X Incorrect; Try Again; 5 attempts remaining Part B - Normal Component of the Reaction at A Determine the normal component of the reaction the rod exerts on the pin at A. Use the coordinate system set up by the free-body diagram below. NO Figure 1 of 1 Express your answer to three significant figures and include the appropriate units. View Available Hint(s) I ? An= Value Units Submit

Answer 1

sof Given that mallkg T=235 kom 82016, 222gom h=620m 023.10 free body diagram Be² hi L2-2hl Cos (90-0) BC = 1 (1-2) 72:95% 2x1.2X2,96%08 190-3-) h B 90-0 = 3:07 m Length of Cable intially Bo= rethy = √ (29+ (1-20) An = 30138 m Applying Sine rule sin (90-0) Sino h BC

(2.9x146870 SX sin (22092) - 21x71x9.81 Cos(32)=7182.9 Sing Sin (90-301) 3.07 1-20 0=2292° are know that Ifr Them Calculate force in cable is F = 2.35X203 0.16 [F=14687.5N taking Moment about A If sing) - (2) w coso =( EmA = Id = ( mla tee 3 15613.40=199,03 you x=78.44 rad's Est = Men At-wcaso + fring=mL z ja At - (71X9.81X60$ (3,1))+(14687.5 Sin(2297))-71(23) 78.44 Az =8009.88 TAL-8.009 KN

Apply Conservation of snergy When 0-30° (k•E +re)intial + Wiz=CkEtr. El final W1-2 = f(BG-BC) =14687.5 (3.138-3.07) =998.75 Nm 070 +993.75= {x (711(209) (29) W4(2.2 -1) (21x2--)) 998.75=99. siw+546) W = 3.08 radys finally nemal Component Efn=now? An-losino - f cosø- m (2) w2 An= -(71x 22x3-08)347Xq.83%8113.))+(146370563627 An=13728.8) | An=13072 KN