Mathematical problem & solution for Force, Stress, Energy

 

Problem:

A cylindrical rod of length $𝐿=2$ meters and cross-sectional area $𝐴=0.01$ square meters is made of a material with Young's modulus $𝐸=200$ GPa. A tensile force $𝐹=10,000$ Newtons is applied along the length of the rod.

1. Calculate the stress in the rod.
2. Calculate the strain in the rod.
3. Calculate the elongation of the rod.
4. Calculate the elastic potential energy stored in the rod due to the applied force.

Solution:

1. Calculate the Stress

Stress ($𝜎$) is defined as the force ($𝐹$) applied per unit area ($𝐴$):

$$𝜎=\frac{𝐹}{𝐴}$$

Given:

• $𝐹=10,000$ N
• $𝐴=0.01$ m²

$$𝜎=\frac{10,000\text{ N}}{0.01{\text{ m}}^2}=1,000,000{\text{ N/m}}^2=1\text{ MPa}$$

So, the stress in the rod is $1\text{ MPa}$.

2. Calculate the Strain

Strain ($𝜖$) is defined as the stress divided by Young's modulus ($𝐸$):

$$𝜖=\frac{𝜎}{𝐸}$$

Given:

• $𝜎=1\text{ MPa}=1\times 10^6{\text{ N/m}}^2$
• $𝐸=200\text{ GPa}=200\times 10^9{\text{ N/m}}^2$

$$𝜖=\frac{1\times 10^6{\text{ N/m}}^2}{200\times 10^9{\text{ N/m}}^2}=5\times 10^{-6}$$

So, the strain in the rod is $5\times 10^{-6}$.

3. Calculate the Elongation

Elongation ($\mathrm{\Delta }𝐿$) is calculated by multiplying the original length ($𝐿$) by the strain ($𝜖$):

$$\mathrm{\Delta }𝐿=𝐿\cdot 𝜖$$

Given:

• $𝐿=2$ m
• $𝜖=5\times 10^{-6}$

$$\mathrm{\Delta }𝐿=2\text{ m}\times 5\times 10^{-6}=10\times 10^{-6}\text{ m}=10\text{ micrometers}$$

So, the elongation of the rod is $10$ micrometers.

4. Calculate the Elastic Potential Energy

The elastic potential energy ($𝑈$) stored in the rod is given by:

$$𝑈=\frac{1}{2}𝜎𝜖𝑉$$

where $𝑉$ is the volume of the rod. The volume $𝑉$ is the product of the cross-sectional area ($𝐴$) and the length ($𝐿$):

$$𝑉=𝐴\cdot 𝐿$$

Given:

• $𝐴=0.01$ m²
• $𝐿=2$ m

$$𝑉=0.01{\text{ m}}^2\times 2\text{ m}=0.02{\text{ m}}^3$$

Now, calculate the energy:

$$𝑈=\frac{1}{2}\times 1\times 10^6{\text{ N/m}}^2\times 5\times 10^{-6}\times 0.02{\text{ m}}^3$$

$$𝑈=\frac{1}{2}\times 1\times 10^6\times 5\times 10^{-6}\times 0.02$$

$$𝑈=\frac{1}{2}\times 5\times 10^{-2}=2.5\times 10^{-2}\text{ J}$$

So, the elastic potential energy stored in the rod is $0.025$ Joules.

Summary:

• Stress in the rod: $1\text{ MPa}$
• Strain in the rod: $5\times 10^{-6}$
• Elongation of the rod: $10\text{ micrometers}$
• Elastic potential energy stored in the rod: $0.025\text{ J}$