A lead brick with the dimensions shown in the fi

Physics
Tutor: None Selected Time limit: 1 Day

Apr 17th, 2015

he force on the brick is 2750 N

This has a vertical component (Fv) of 2750 * Sin(21)
= 985.5 N

and a horizontal component (Fh) of 2750 * Cos(21)
= 2567.3 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 

= Fv * h / (A * dh)

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 = 985.5 * 0.02 / (0.003 * dh)

dh = 985.5 * 0.02 / (0.003 * 16 * 10^9)

= 4.106 * 10^-7 m

(b)
Shear modulus for Lead = 5.6 GPa

= 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 = 2567.3 * 0.02 / (0.003 * dx)

dx = 2567.3 * 0.02 / (0.003 * 5.6 * 10^9)
=
= 3.056 * 10^-6 m

Apr 17th, 2015

part b is incorrect

Apr 17th, 2015

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Apr 17th, 2015
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Apr 17th, 2015
Dec 9th, 2016
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