This has a vertical component (Fv) of 2050 * Sin(27) = 930.68 N and a horizontal component (Fh) of 2050 * Cos(27) = 1826.56 N We have to assume that the lead brick will undergo elastic deformation rather than plastic deformation. (a) Young's modulus for Lead is 16 GPa (from Wikipedia) = Fv * h / (A * dh) (from Wikipedia) where h is the height = 0.02 m A is the top area = 0.06 * 0.05 = 0.003 m^2 dh = the change in height 16 * 10^9 = 930.68 * 0.02 / (0.003 * dh) dh = 930.68 * 0.02 / (0.003 * 16 * 10^9) = 387.78 * 10^-9 m = 3.878 * 10^-7 m (b) Shear modulus for Lead = 5.6 GPa (from Wikipedia) = Fh * h / (A * dx) where dx is the amount the top surface will move horizontally because of the defromation (from Wikipedia) 5.6 * 10^9 = 1826.56 * 0.02 / (0.003 * dx) dx = 1826.56 * 0.02 / (0.003 * 5.6 * 10^9) = 2174.5 * 10^-9 m = 2.175 * 10^-6 m

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