368_Physics ProblemsTechnical Physics

368_Physics ProblemsTechnical Physics - 370 P12.52 Static...

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370 Static Equilibrium and Elasticity P12.52 (a) Just three forces act on the rod: forces perpendicular to the sides of the trough at A and B, and its weight. The lines of action of A and B will intersect at a point above the rod. They will have no torque about this point. The rod’s weight will cause a torque about the point of intersection as in Figure 12.52(a), and the rod will not be in equilibrium unless the center of the rod lies vertically below the intersection point, as in Figure 12.52(b). All three forces must be concurrent. Then the line of action of the weight is a diagonal of the rectangle formed by the trough and the normal forces, and the rod’s center of gravity is vertically above the bottom of the trough. A B F g O FIG. P12.52(a) (b) In Figure (b), AO BO cos . cos . 30 0 60 0 °= ° and L LL 2 22 2 2 30 0 60 0 30 0 60 0 1 2 2 2 =+=+ ° ° F H G I K J = + = ° ° AO BO AO AO AO cos . cos . cos . cos . So cos θ == AO L 1 2 and 60 0 .. A B F g O 30.0 ° 60.0 ° FIG. P12.52(b) P12.53 (a)
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