##### A 0.9 kg block of ice is initially at a temperature of -5°C.

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Apr 16th, 2015

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

Apr 16th, 2015

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Apr 16th, 2015
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Apr 16th, 2015
Dec 8th, 2016
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