details FULI SCREEN PRINTER VERSION BACK Chapter 12, Problem 031 Your answer is
ID: 1781070 • Letter: D
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FULI SCREEN PRINTER VERSION BACK Chapter 12, Problem 031 Your answer is partially correct. Try again In the figure, a nonuniform bar is suspended at rest in a horizontal position by two m makes the angle 67.2" with assless cords as shown in the figure here. One cord bar is 2.6 m, compute the distance x from the left end of the bar to its center of mass makes the angle 9-22.a with the vertical; the other the vertical. If the length L of the com 0.78 Sons.Inc. All Rights Reserved. A Division of John Wiley&Sons;,Inc.Explanation / Answer
along horizontal
Fnet,x = 0
-TL*sintheta + TR*sinphi = 0
TL = TR*sinphi/sintheta
TL = TR*sin67.2/sin22.8
TL = 2.38*TR
along vertical
TL*costheta + TR*cosphi = W
2.38*TR*cos22.8 + TR*cos67.5 = W
TR = W/(2.38*cos22.8 + cos67.5)
TR = 0.39*W
TL = 0.93*W
Net torque about the left end of the rod = 0
TR*cosphi*L - W*x = 0
0.39*W*cos67.2*2.6 - W*x = 0
0.39*cos67.2*2.6 - x = 0
x = 0.393 m <<<<<------------ANSWER
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