A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2 . It is ...
from the Hooke’s law :
ε = σ/Y = (1.8 x 10^8 Pa)/(20 x 10^10 Pa)
= 9.0 x 10^−4
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2 . It is ...
To determine the resulting strain, we need to calculate the stress first.
The stress (σ) is given by the formula:
σ = F / A
Where:
σ = stress
F = force applied
A = cross-sectional area
In this case, the force applied is the weight of the milling machine, which is given by:
F = m * g
Where:
m = mass of the milling machine
g = acceleration due to gravity
In this case, the mass of the milling machine is 550 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging in these values, we can calculate the force:
F = 550 kg * 9.8 m/s^2
F = 5390 N
Now, we can calculate the stress:
σ = 5390 N / 0.30 cm^2
Since the cross-sectional area is given in cm^2, we need to convert it to m^2:
1 cm^2 = 0.0001 m^2
Therefore, the cross-sectional area in m^2 is:
0.30 cm^2 * 0.0001 m^2/cm^2
0.00003 m^2
Plugging in the values, we can calculate the stress:
σ = 5390 N / 0.00003 m^2
σ = 179,666,666.67 N/m^2
Now that we have the stress, we can determine the strain using Young's modulus (Y). The formula for strain (ε) is:
ε = σ / Y
Plugging in the values, we can calculate the strain:
ε = 179,666,666.67 N/m^2 / 20
ε = 8,983,333.33
Therefore, the resulting strain is approximately 8,983,333.33.