Question

Question 5 The two 14-kg blocks are connected by a light rod, and the assembly lies in a vertical plane. The system is released from rest with 0 = 50°. (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 O at B, what is the minimum value of us at A for static equilibrium? B 69 А Answers: (a) Hs = (b) Hs =

Answer 1

A detailed solution is given in the image. Hope it's helpful.

Allt SOLUTION: 8 Naho 210 IR 690 B For the System to be in equilibrium taking moment about o for equilibrium. [42] EMO 0-500 0*7690 A on 'fa e INA mA = mB = 14 kg NA-Ng = 2*14*9.81 = 206.01 N vo -> 0 NA = Ng + 206.01(N) [#] Egy = 0 - For + Ng Cos 21° + Focos 69° = 0 fa fB but FFA = MSNA ; ffo = Mg No =) - USNA + 0.934 Mo + 0.36 fps = 0 => -lls ( Ng+206.01) + 0.934 Ng + 0.36 MsNo [Substituting o] - elg (Ng+ 206.01 -0.36 No) +0.934 Ng 0.934 Ng .: Ms = 0.64 Ng+206.01 (tm) Efy = 0 NA - MAg-mg9 – Ng Sin 21° + F Sin 69° - 0 fo Ng+ + 206:ol - (2= lq +4.8) - 0:36 Na + 1, Na+ 0:34) = 0 0.64 NB 68.58 + 0.934 lg Ng Ms 68.58 -0.64 NB 0.934 Ng

egyreating ☺ & © 0.934 No 68.58 - 0-64 NB 0-934 NB 0.64 Ng+ 206.01 → (0.934 Ng) = (6858 *0.64 NJ+(68.58 * 206.01) -(0.64 Ng)2-(0-642206-01 Ng) => 0-872356 Ne 43:89 Na + Iq128.166 - 0:406 N. - 3 8464 Na 1-281956 Ng? + 87.9564 No - 14128.166 Solving for No we get No = 76.14 N :: No = 282.15 N ولا .: 0.934 * 76.14 (0.64*76-14 + 206.01) ولل = 0.279 ANSWER (Ь In this problem, ffe =0 :: (FEMO=0 =>(magul) - [Ng*f] + (m99*)+(gul) = NA = NB + 206.01 (7) Efa - FFA + Ng CoS 210 =0 >> -Us Not 0:934 Ng = 0 [ substitute o 0] =) - Us ( Ng+ 206.01) +0.934 NB 2 i. Ms 0.934 Ng Na + 206.01 (+) { Fy - Ng Sin 210 = 0 NA-MAg - mag 0.36 Ng NB +206.01 2*14*9.81 107. 297 N - © :. Ng

Substitute 3 in a 0.934 * 107. 297 lls = -32. 107• 297 + 206.01 ولد .:: = 0.32 ANSWER