A 3700 kg statue is placed on top of a cylindrical concrete (Y = 2.3 1010 N/m2) stand. The stand has a cross-sectional area of 7.2 10-2 m2 and a height of 2.0 m. By how much does the statue compress the stand?
They give you Young's moduls of the concrete, Y = 2.3*10^10 N/m^2
The pedestal area is A = 7.2*10^-2 m^2
Use the elastic law that
stress = (strain)*Y
In this case,
strain = compression/length
Therefore
compression = Length*strain
= Length*stress/Y
= Length *(weight)/[(area)*Y]
Don't forget that weight = M g, so
compression = L M g/[(area)*Y]
To find out how much the statue compresses the stand, we need to apply the concept of pressure and use Hooke's law. Hooke's law relates the amount of compression or extension of an elastic material to the force applied to it.
First, let's calculate the force exerted by the statue on the stand. We can use the formula:
Force (F) = mass (m) × acceleration due to gravity (g)
Here, the mass of the statue is given as 3700 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore,
F = 3700 kg × 9.8 m/s^2 = 36260 N
Now, let's calculate the pressure exerted by the statue on the stand:
Pressure (P) = Force (F) / Area (A)
Here, the area of the stand is given as 7.2 × 10^-2 m^2. Thus,
P = 36260 N / (7.2 × 10^-2 m^2) = 503611 N/m^2
Next, let's calculate the amount of compression (Δh) using Hooke's law:
Δh = (F / A) / Y
Here, Y is the Young's modulus of the material, which is given as 2.3 × 10^10 N/m^2. Therefore,
Δh = (36260 N / (7.2 × 10^-2 m^2)) / (2.3 × 10^10 N/m^2)
Simplifying the above expression, we get:
Δh ≈ 0.079 meters or 79 mm
Therefore, the statue compresses the stand by approximately 79 mm.