Question

The two 10-kg blocks are connected by a light rod, and the assembly lies in a vertical plane. The system is released from rest with 0 = 67º. (a) Determine the minimum value of the coefficient of static friction common to both A and B for which the system will remain in static equilibrium. (b) If us = 0 at B, what is the minimum value of us at A for static equilibrium? 58° BA Answers: (a) us = (b) us =

Answer 1

(A) 0.310

(B) 0.351

Y. NB 32° txe را. - (0) 500 mg (a.) we must assume that slippling occurs at A and B simultaneously and FA : Ms NA Fa: lls. No Now, we have 3 unknowns, CNA, NB & g). Equating som of moments about point o EMO=0 No. e a & mg 2 LEA 0 mg to zero , NA A. ½ 42 mg direction to zero we +mg. I - Nail=0 e = length between O and A NB & mg 1 + mg-NA=0 of O and B. NA = No + 105 mg = NA + 1.5 X 10 X 9.81 .. NA = No + 147.15 (ir). By equating sum of forces in se can write us as fonction of No. EF=0. -FA + NA LOS 32° -FA + 0.848 Ng+ 0.530 Fo=0 Using equaliou land Gi'), -( NA Us I +0:848 NB +0.530F8-0 :-(Ng +147.15).s +0:848 NB +0.530 No es = 0 3-ills (Ng +147.15 - 0:530 Ng) + 0.848 Ne=0 58° - 0 + FB LOS 0.848 No (0 es - 6:53 Ô Ma + 141 IC

- Finally, by equaling sum of forces in y-direction to zero, we can determine force NB {Fy=0 sin 5°=0 NA-ling-Ng sin 32% + F NA -2.10.9.81 - 0.530 NA +0.848 FB = 0 NA - 196.2 - 0.53 NB +0.848 Fo=0 Replace FB and NA from equation is and (ii), NB + 147.15 196.2 -0.53 NB +0.848 No els = 0 0.47 No - 49.05 +0.848 Ngills 30 0.848 No. els : 49.05 - 0:47 NB Ms 49.05-0.47 NB 5) 0.848 NB NOW, compare equation and , 0.848 No 49.05 - 0.47 MB 0.53 NB +147:15 0.848 NB 0.719 Ng -0.249 Ng? -43.164 Ng + 7217.7 0.968 Ng + 43.164 NB 7217.7 - 0 solving the quaratic equation, we get 2 2 NB: 66.88 Ni Finally, 56.714 182.59 0.848 x 66.88 0.53 X 66.88 +147.15 M Us = 0.310

(b) NE since FB=0. and FA = ls. NA ng l ing 58 -FA NA - FA+Ng Cos 32=0 Sum of moments about 0, EM₂-0 NB l + mg 1/2 ting. I-NA-1=0 NA - NB + 147.15 Equale force in x-direcion. {Fx=0 -FA +0.848 NB=0 Replace A with eGNA. Naus +0.848 Ng=0 Reblace NA by expression ③ (NB+147.15 Jels +0.848 Ne=0 0.848 NB us NB + 147.15 Similarly in ydirection - Efy=0 NA-ang-Ngsin32 = 0 NA – 196.2 -0.53 NB=0 NB + 41: - 1362 -0.53 MB = 0 NBC 49.05 = 104.36 N. 0.47 put this in expression 0.848 X 104:36 Us : = 0.351 104.36 + 147.15 dg 035 An. 4 = 0.351