43. (a) The shear stress is given by F/A, where Fis the magnitude of the force applied parallel to one face of the aluminum rod and Ais the cross–sectional area of the rod. In this case Fis the weight of the object hung on the end: F= mg, where mis the mass of the object. If ris the radius of the rod then A= πr2. Thus, the shear stress is
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.