Question

II Review The forked rod is used to move the smooth 3-lb particle around the horizontal path in the shape of a limaçon, r = (2 + cos 0) ft.(Figure 1) Part A If 0 = (0.5t2) rad, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the particle on only one side. Express your answer with the appropriate units. Figure 1 of 1 μΑ ? F = Value lb Submit Previous Answers Request Answer 3 ft X Incorrect; Try Again; 5 attempts remaining

Answer 1

full body diagram N fe, tangent as F 3ft at 3 lb particle, normal and tangential force exist shown They are not aligned since the shape of path is not perfect circde. Given: r= 2 + cost differenhale &=-8. sine ② -å sina 8² cose 2 Given: 0 = 0.5t2 å = 2x0.5 t = t o 1 - Oradiso مع Now at the instant of 1 second angle o from above o equations. e = 0.5812²= 0.5 rad I radis ös 1 rad/s2 and put o&o in above é equations: r = 2 + cos 0.5 2.877 ft B- a radians e = -1* sin(0.5) -1.357f4152 Now: radial á (1), sin 0.5 = -0.479@ft/s 1² cos (o.s) = and tangential acceleration ar - i - re2 = -1.357 - 2.877x 12 = -4,234 ft/52 ao - ró +240 = 2.877 *1 + 2x(-0,479) (1) - CHW+ 2 vw) - 1.919 ft/s

angle a tana r 2+coso - sino S: drome de sino JO deydo at a=0.5 rad tan a 9,46° 2+ cos 0.5 = -6.00214 - Sinos d = -1.405 kad, -1,405 X 180 = -80.5410 a ongle: = go-80.541 Equilibuirem of forces: along radial direction : E re- 0 m - NcOS 9.46 = 2 (-4,234) 32.2 N- 0.266 lb along o' direction : E Fo=mages; N sing. 46° = 2 x 1.919 90 ) K LL 32.2 f = 0.163 lb E