Question

A uniform wooden bar (M = 6.81 kg, L = 51.2 cm) is free to rotate about a frictionless axis in the center. If forceF₁ = 40 N acts at angle θ₁ = 67^o at distance L/3 from the left end, and force F₂ = 89.1 N acting at angle θ₂ = 59.8^o on the right end, calculate α, the angular acceleration of the bar, in rad/s². The direction is given by the sign: positive if it is counter-clockwise, and negative if it is clockwise.(refer ti diagram 2

HW Diagrams and Extra Problem x Take Test: Physics 121 Spring 202 X 3 Microsoft Word - physics 121 dia x C Gat Homework Help With Chegg x + - 3 x E c O File C:/Users/andre/Downloads/physics%20121%20diagramsikproblems%20B(1).pdf DIAGRAMS FOR ASSIGNMENT 8 Diagram 3 Diagram 1 Diagram 2 Axis L3 View from above Diagram 4 Diagram 5 Diagram 6 Axis Axis - View from above View from above Type here to search O EM e 6 W A.ORG ENG 211/2006

Answer 1

SEC */ Axuj Élisata Sint LA 2 Esone Given Values: Mass Me 6.81kg, L: 51.2cm=0.512 m. Fi = 40N , 0,= 67°, F2 = 89.1 N 02 = 59.8° da? lorque due to componen) of foree fi Cifra Fisites Fisinoofb-) - Fisinau = hosinty".0.512) N-m = +hify anticlockwise = = Filoso, 319 N-m Filoso, y 8t fcaso, XO = O Nam Torque due to component of fore e fa sino - Ę Snox = 89.1 Sim 59.80 x 0.512 --191713 Cloc 2 . Clock wise . - Floso Coso xo Net torque Enet - Casino, +4, 600, ter sine to come = 3.13 +0.t-19.313.70 - = -16.583 - sign shows clockwise

T= I LA - 1657 = M1² a. Irod, = M22 (about com). -16:57 = 6.81*(0,512) «. 12 12 Q = -16.57x12 6.81X0.5 122 a = - 111.38 rad/s2. llal = 111038 rad) s4 Direction & Clockwise)