00 m measured rightward from the hinge and at x2 300 m

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Unformatted text preview: and ℓ as the length of the boom, we compute torques about the hinge and find l Mg l sin β + mg ( 2 ) sin β Mg sin β + mg sin β / 2 T= = . l sin α sin α The unknown length ℓ cancels out and we obtain T = 6.63 × 103 N. (b) Since the cable is at 30º from horizontal, then horizontal equilibrium of forces requires that the horizontal hinge force be Fx = T cos 30° = 5.74 ×103 N. (c) And vertical equilibrium of forces gives the vertical hinge force component: Fy = Mg + mg + T sin 30° = 5.96 ×103 N. 12P30. (a) The sign is attached in two places: at x1 = 1.00 m (measured rightward from the hinge) and at x2 = 3.00 m. We assume the downward force due to the sign’s weight is equal at these two attachment points, each being half the sign’s weight of mg. The...
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