A metal bar with a cross-sectional area of 50mm2 is subjected to a tensile strain of 0.001. What is the tensile force responsible for the strain given that the material has a Young's modulus of 15GNm raise to the power -2
A 0.003N
B 7.5x103N
C 7.5x 10-10N
D 3x1012N
To find the tensile force responsible for the given strain, you can use Hooke's Law, which states that the stress (force per unit area) applied to a material is directly proportional to the strain (change in length per unit length) produced, assuming the material is within its elastic limit.
Hooke's Law can be expressed as:
Stress = Young's Modulus * Strain
Here, the cross-sectional area (A) of the bar is given as 50 mm^2, which is equivalent to 50 x 10^-6 m^2 (since 1 mm^2 = 10^-6 m^2). The strain (ε) is given as 0.001, and the Young's modulus (Y) is given as 15 GN/m^2 (GigaNewtons per square meter), which is equivalent to 15 x 10^9 N/m^2.
First, let's convert the strain from millimeters (mm) to meters (m):
ε = 0.001 = 0.001 x 10^-3 = 10^-6 m/m (since 1 mm = 10^-3 m)
Now, we can calculate the stress (σ) using Hooke's Law:
Stress = Young's Modulus * Strain
σ = Y * ε
σ = (15 x 10^9 N/m^2) * (10^-6 m/m)
σ = 15 N/m^2
Finally, we can calculate the tensile force (F) responsible for the strain by multiplying the stress by the cross-sectional area:
F = Stress * Area
F = 15 N/m^2 * (50 x 10^-6 m^2)
F = 15 N/m^2 * 50 x 10^-6 m^2
F = 0.75 N
Therefore, the tensile force responsible for the given strain is 0.75 N.
The correct answer choice is A) 0.75 N.