25. The beam is in equilibrium: the sum of the forces and the sum of the torques acting on it each vanish. As shown in the figure, the beam makes an angle of 60º with the vertical and the wire makes an angle of 30º with the vertical. (a) We calculate the torques around the hinge. Their sum is TLsin 30º – W(L/2) sin 60º = 0. HereWis the force of gravity acting at the center of the beam, and Tis the tension force of the wire. We solve for the tension: ()222N sin 60sin60===192 N.2sin30WT°°°°(b) Let Fhbe the horizontal component of the force exerted by the hinge and take it to be
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