Question

End A of the 6-kg uniform rod AB of length L = 1.5 m rests on the inclined surface, while end B is attached to a collar of negligible mass which can slide along the vertical rod shown. When the rod is at rest a vertical force P is applied at B. causing end of the rod to start moving upward with an acceleration of 4 m/s2 Knowing that = 35°, determine the force P. В The force Pis N1.

Answer 1

The force P is 59.5 N

solution i kinemabis : AB= 4 m/s AB= 4 m/² B 3 0.15 QA= Cipta AB 125 = 4h +105x 35 A aa law of sines : RA/B=1.54 4 1.5 Si65 sis 60° 60 204 x = 2.7967 rad/ 65 CA a = AG = AB+ a Alb 47 +0,75 (1.7907) Da ax = 0,15 (2.7967) to 335° ax 1.7145 m/

ay = 47 + 0.45 (207904) sis 350 L ③ ay = 2:7995 m/s2 1 kinetics : me² 6 kg (1.5m² I- 12 I = 1.105 kg me B. : 8 559 0.75m max พ may A A Law 여 sines : AB EB sisbo sis 65 2 EB = sis bo (15) bistro 1. 4333 m = FB t ED = EB-OB 1.4333 -0.75 605 55 ED = 1.0031 m. DG = 0.75 is si 55° 0.61436 m = DG

3 + ) + ZME = = Me) en eft p(EB) - WCED) - IX tmax (DG) + may(ED) (1.4333) – 6(9.8) (1,0031) = 1,125(2.7904) + 6(1.9145) (0.6143!) + 6(2.7995) (1.0031) 1.4333Þ - 59.0425 = 3.1395 + 6.3199 +16.8491 85.53) 1.4333 p = p = 59.5NT *