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ile Edit View History Bookmarks Develop Window Help WileyPLUS edugen.wileyplus.com/edugen/student/mainfr.uni Wiley PLUS: My wiley PLUS I Help I contact us Log o Cutnell, Physics, 9e GENERAL PHYSICS 1 AND 2 (PHYS 209/210 Assignment Gradebook ment EMY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT Chapter 09, Problem 10 A rotational axis is directed perpendicular to the plane of a square and is located as shown in the figure. Two forces, F 1 and F 2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the figure. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of F2 is 2.5 that of F 1. Find the distances (a) a and (b) b that locate the axis. times (a) Number Units (b) Number Units Click if you would like to show Work for this question: Open Show WorkExplanation / Answer
Net torque = torque due to F1 + torque due to F2 = 0
torque = r X F
due to F1 = b*F1
due to F2 = a*F2 direction opposite of due to f1
if net torque produced is zero then bF1 = aF2
and F2 = 2.5 F1
bF1 = a(2.5F1)
b = 2.5a ........ (i)
in figure b ,
torque due to F1 = (1-a)*F1
due to f2 = bF2 = b(2.5F1)
so, (1-a)F1 = 2.5bF1
2.5b + a = 1 .... (ii)
solving (i) and (ii) ,
we get a = 4/29 = 0.138 meter
b =10/29 = 0.345 meter
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