University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 11.13: In both cases, the tension in the vertical cable is the weight . a) Denote the length of the horizontal part of the cable by L. Taking torques about the pivot point, TL tan 30.0 = wL + w( L 2), from which T = 2.60 w. The pivot exerts an upward vertical force of 2 w and a horizontal force of 2.60 w , so the magnitude of this force is 3.28w , directed 37.6 from the horizontal. b) Denote the length of the strut by L , and note that the angle between the diagonal part of the cable and the strut is 15.0. Taking torques about the pivot point, TL sin 15.0 = wL sin 45.0 + ( w 2) L sin 45, so T = 4.10 w. The horizontal force exerted by the pivot on the strut is then T cos 30.0 = 3.55 and the vertical force is (2 w) + T sin 30 = 4.05w, for a magnitude of 5.38w, directed 48.8. ...
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