Question

Question 12 (1 points) Two wheels A and B with weights w and 2w, respectively, are connected by a uniform rod with weight w/2, as shown below. The wheels are free to roll on the sloped surfaces. Determine the angle that the rod forms with the horizontal when the system is in equilibrium. Hint: There are five forces acting on the rod, which is two weights of the wheels, two normal reaction forces at points where the wheels make contacts with the wedge, and the weight of the rod. Image size: S M L Max - TO 60° 300 DA

Answer 1

. NA = Normal force on A MB = Normal force on B. . 6 - 300 NB Law Free body diagram of syedem is CO tromelation costo aw system is in equilibrium (both nal and rotatatanel). Along Horizontal Med force is zero so, No caso » N3X 1/2 = NA LA - No = BNA Along vertical (cs Scaphed in force is z ero so, CamScanner

No singoº + Na singoº = 2W + + >No V + NAxlL = 70 3. NA TV + = 70 » 4N - 70 N = & W] -:|x6 = w .: system is in equilibraum so, net forque about any point will be zero. 2. Net torque about centre of sphere B is zero. ritmet Tew + Tw tow to o 20+0+149x cosa +(LXCOSO)XW toralsin (309_01_0 Because angle between rod and Na ces(3820) .; y caso + WEcoso = Zbotsin 68-6) > Ecoso = sen (88–0) → 50050= 7 senzocoso C333093em o7 cs sempes cose = 76. cola - 3 sino

cosa G >= ? Commen were >1=7 [- tana] 915 7o - The smo → 73 tema = 24 = 440-- > tamo = BO > 102-13.90) e ve sign as indicating that sphere A will be below herezontal and sphere B will be above horizontal by angle 13.go which we have taken in figure en opposite way. . - Scanned with CamScanner