4. 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 7.5x10 3N
C 7.5x 10 -10N
D 3x10 12N
To find the tensile force responsible for the given strain, we can use Hooke's Law formula. Hooke's Law states that stress (force per unit area) is directly proportional to strain (change in length per original length) within the elastic limit of a material.
The formula for stress is: Stress = Young's modulus * Strain * Area
In this case, the strain is given as 0.001, the Young's modulus is 15 GN/m^2, and the cross-sectional area is 50 mm^2.
First, we need to convert the area to square meters:
Area = 50 mm^2 * (1 m^2/1,000,000 mm^2) = 0.00005 m^2
Now we can calculate the tensile force using the formula:
Stress = Young's modulus * Strain * Area
Stress = (15 GN/m^2) * (0.001) * (0.00005 m^2)
Note: GN stands for gigan