Question

The elements of a switching device are shown. If the vertical control rod has a downward velocity v = 3.2 ft/sec and an upward acceleration a = 1.9 ft/sec2 when the device is in the position shown, determine the magnitude of the acceleration of point A. Roller C is in continuous contact with the inclined surface. A04" 350 18 87.8284 Answer: aA = ft/sec2

Answer 1

Ve= -3.2j - JE JB + WBC X ABC ve [ comisi + SM 185] = -3.2; + Wack* [ 4.com 35. 72 ve { coolgi + 5n18;] = 011912 woc i + (0-27307 W8<-3-2) Equating t e j coefficients. 0.1912 Woc - a 0.951 Vc = 0-309 Vc = 0.27277 Woc – 3:2 - Solving this, we get . - 15.171 radis. Now, ae = abt Tact Tec - were no ac [colli + Sint&j] = 1.9 ; + 4Bc* [ 4. ces35i - 4 SM3sj] - 15.1715 [Conssia Sin35j] Qe [col8i + sm 18;] = (0.191208c- 62-845) i +(0.273 dBc + 45.9);

Equating in coefficients, we get 0.951 ac = 0.1912 066 - 62.845 8 0.309 ac = 0.273 ABC + 45.9 solving these equation, we get . = 120.8 rad's ? Now, LAB= a Be a WAB=WBL (same link) TA = TB + TABX LAB - WAS LAB I ::9A= 1.9; + 120.8 k*[-. cosas it 4.61 35; 7 - (15.cana [-4 assisi + 46M25j7 ? 12 ] 4A = 39.75 i - Yi PAU S 39.75°+ 75:12 1941 = 84.96 61132 / Am