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Question
IA University of South Alabama MA227 y WileyPLUS Ahmad > c Secure l https://edugen.wileyplus.com/edugen/student/mainfr.uni Bing w Wikipedia R Facebook y Twitter-LinkedIn D The weather Channel Yelp »Other Bookmarks I Help I Contact Us I Leg Out ps WileyPLUS Differential calculus. LA University of South D Apple iCloud Yahoo WileyPLUS Halliday, Fundamentals of Physics, 10e Calculus-based Physics I ⅈ (PH 201-202) Home Read, Study & Practice Gradebook ORION Downloadable eTextbook Assianment> Open Assignment FULL SCREEN PRINTER VERSION BACK ASSIGNMENT RESOURCES Chapter 10, Problem 046 201-14 The body in the figure is pivoted at O. Three forces act on it in the directions shown: FA = 10.0 N at point A, 6.30 m from 0; FB = 19.0 N at point B, 2.10 m from 0; and FC = 20.0 N at point C, 3.70 m from O. Taking the clockwise direction to be negative, what is the net torque about 0? 034 048 050 054 0 Number Units the tolerance is +/-2% LINK TO TEXT Question Attempts: 0 of 6 used SAVE FOR LATER SUBMIT ANSWER Version 4.24.1.23Explanation / Answer
torque = r*F*sintheta
theta = angle between r and F
torque due to FA = FA*OA*sin(180-135) = 10*6.3*sin(180-135) = 44.54 Nm
torque due to FB = FB*OB*sin(90) = 19*2.1*sin(90) = -39.9 Nm
torque due to FC = FC*OC*sin(180-100) = 20*3.7*sin(80) = 72.9 Nm
total torque = +77.54 Nm