Question

Problem 13.95 A smooth can C, having a mass of 3.5 kg. is litted from a feed at A to a ramp at B by a rotating rod. The rod maintains a constant angular velocity of 0 - 0.5 rad/s, Neglect the effects of friction in the calculation and the size of the can so that r = (1.2 cos 0) m. The ramp from A to B is circular, having a radius of 600 mm. (Figure 1) Part A Determine the magnitude of the force which the rod exerts on the can at the instant 8 - 30° Express your answer to three significant figures and include the appropriate units. F- 19.02 N Submit Previous Answers Beguest Answer Figure < 1 of 1 > * Incorrect; Try Again; 11 attempts remaining Review your calculations and make sure you round to 3 significant figures in the last step, B Provide Feedback 0=0.5 rad/s 600 mm le 600 mm

Answer 1

Ans: Given: m = 3.5kg ġ = 0.5 Had/sec. (1-2 cosom Equilibrium equation in the 7 Efr=mar cylindarical coordinates m (r-rón) ==0 = m Crö +2 ro) T EFE mao r = 1.2 coso. ř = -102 sino.o. r = -2 TOSO. 02- 1:2.sino. for o= 30°. 6 = 0.5 rad/s. Ö = 0. r = 1.2. C0830 1.0392 m. r = -1.2x Sinsox 0-5= -0.3 m/s. = -1.2 x Cos 38 0.5?- 1-2x8in 3oxo = - 0. 2598 mis-

..Substitute the values in equation 0 cmd @ Efr = Cr-r8² 3-5(-0. 0-254€ - (1-0392) (0.5))). - ..Efr=-108186.N +2 (-0.3) (0-5)) {FO = m (rö + 2ř ) = 3-5 (C1-0392) (0) -EFO =-1.05N from the FBO : mg /Y 300 R. 30° mg sino Efr = N coso - mg -lo 686 No Cas 300 – 6-5) (9-81). Sinzo N = 17.72 N. i SFO - Ft Nsino mg coso. - 1:05 = Ft 17.72 sino - (3-5) (9.8). Cop300. -1.05 = 8.86 -29.735 1 F = 19.825 N S