How Much Pressure Is Required to Compress an Iron Block by 0.16%?

[zyphriss2]

How much pressure is needed to compress the volume of an iron block by 0.16 \%? Express your answer in {\rm{N/m}}^2.

I know pressure is the same as stress, and that stress is force/csa. Could someone point me in the right direction.

 

[Redbelly98]

Look up the relation between stress and strain.

Also, you will need the value of Young's modulus for iron.

 

[thomate1]

The pressure on an iron block can be calculated using the formula P = F/A, where P is pressure, F is force, and A is the cross-sectional area of the iron block. In order to compress the volume of the iron block by 0.16%, we can use the formula V2 = V1(1-ε), where V2 is the final volume, V1 is the initial volume, and ε is the strain (0.16%).

First, we need to determine the force required to compress the iron block. This can be done by using the formula F = kΔL, where k is the bulk modulus of iron and ΔL is the change in length due to compression. The bulk modulus of iron is approximately 170 GPa (gigapascals).

Next, we need to determine the cross-sectional area of the iron block. This can be calculated by dividing the volume by the length, as A = V/L.

Putting all these values together, we get:

P = (kΔL)/(V/L)

Substituting in the values for k, ΔL, and V, we get:

P = (170 GPa)(0.0016)(1 m^3)/(1 m^2)(1 m)

P = 272,000 N/m^2

Therefore, the pressure needed to compress the volume of an iron block by 0.16% is approximately 272,000 N/m^2. It is important to note that this is an approximate value and may vary depending on the specific properties of the iron block.