A lead brick with the dimensions shown in the figure below (6.0 cm left to right

Physics
Tutor: None Selected Time limit: 1 Day

Apr 16th, 2015

The force on the brick is 2750N
This has a vertical component (Fv) of 2750 * Sin(27) 
= 1248.47 N 

and a horizontal component (Fh) of 2750 * Cos(27) 
= 2450.27 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 = 1248.47 * 0.02 / (0.003 * dh) 

dh = 1248.47 * 0.02 / (0.003 * 16 * 10^9) 
= 5.2*10^(-7)

(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 = 2450.27 * 0.02 / (0.003 * dx) 

dx = 2450.27 * 0.02 / (0.003 * 5.6 * 10^9) 
= 2.92 *19^-6

Apr 16th, 2015

the answers were incorrect


Apr 16th, 2015

The force on the brick is 2750N
This has a vertical component (Fv) of 2750 * Sin(21)
= 985.51 N 

and a horizontal component (Fh) of 2750 * Cos(21)
= 2567.35 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 = 985.51 0.02 / (0.003 * dh) 

dh = 985.51 0.02 / (0.003 * 16 * 10^9) 
= 4.11*10^-7

(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 = 2567.35 0.02 / (0.003 * dx) 

dx = 2567.35 0.02 / (0.003 * 5.6 * 10^9) 
= 3.1*10^-6

Apr 16th, 2015

it is still incorrect :(

Apr 16th, 2015

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