370 Static Equilibrium and ElasticityP12.52(a)Just three forces act on the rod: forces perpendicular to thesides of the trough at A and B, and its weight. The lines ofaction of A and B will intersect at a point above the rod.They will have no torque about this point. The rod’s weightwill cause a torque about the point of intersection as inFigure 12.52(a), and the rod will not be in equilibriumunless the center of the rod lies vertically below theintersection point, as in Figure 12.52(b). All three forcesmust be concurrent. Then the line of action of the weight isa diagonal of the rectangle formed by the trough and thenormal forces, and the rod’s center of gravity is verticallyabove the bottom of the trough.ABFgOFIG. P12.52(a)(b)In Figure (b), AOBOcos.cos.30 060 0°=°andLLL2222230 060 030 060 01222=+=+°°FHGIKJ=+=°°AOBOAOAOAOcos.cos.cos.cos.So cosθ==AOL12and =°60 0..ABFgO30.0°60.0°FIG. P12.52(b)P12.53(a)
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