Determine the elongation of the rod in if it is under a tension of 5.7E10^3 N.
The rod is made of aluminum and copper. From the aluminum to the copper section it measures 1.3m and from the copper section to the end it measures 2.6m with a radius of 0.2cm.
Can someone please help, I honestly don't know where to begin.
duplicate post; already answered
Sorry about that!
To determine the elongation of the rod under tension, we can use the formula for strain, which is the ratio of the change in length to the original length:
ε = ΔL / L
where ε is the strain, ΔL is the change in length, and L is the original length.
First, let's calculate the change in length separately for the aluminum and copper sections.
For the aluminum section:
Since we know the tension force, we can use Hooke's Law, which states that stress is proportional to strain:
σ = E × ε
where σ is the stress, E is the modulus of elasticity, and ε is the strain.
The modulus of elasticity for aluminum is approximately 70 GPa (GigaPascals) or 70 × 10^9 Pa.
We can rearrange the equation to solve for strain:
ε = σ / E
Substituting the given tension force:
ε = (5.7 × 10^3 N) / (70 × 10^9 Pa)
ε = 8.14 × 10^-8
To find the change in length, we multiply the strain by the original length:
ΔL_aluminum = ε × L
ΔL_aluminum = (8.14 × 10^-8) × 1.3 m
ΔL_aluminum = 1.06 × 10^-7 m
For the copper section:
The modulus of elasticity for copper is approximately 117 GPa (GigaPascals) or 117 × 10^9 Pa.
Using the same formula for strain:
ε = σ / E
ε = (5.7 × 10^3 N) / (117 × 10^9 Pa)
ε = 4.87 × 10^-8
To find the change in length for the copper section:
ΔL_copper = ε × L
ΔL_copper = (4.87 × 10^-8) × 2.6 m
ΔL_copper = 1.26 × 10^-7 m
Now, to calculate the total elongation of the rod:
ΔL_total = ΔL_aluminum + ΔL_copper
ΔL_total = 1.06 × 10^-7 m + 1.26 × 10^-7 m
ΔL_total = 2.32 × 10^-7 m
Therefore, the elongation of the rod under a tension of 5.7 × 10^3 N is approximately 2.32 × 10^-7 m.
To determine the elongation of the rod, we can use the formula for axial strain:
ε = ΔL / L
where ε is the strain, ΔL is the change in length, and L is the initial length of the rod.
To find the change in length, we need to consider the properties of both aluminum and copper and their respective sections of the rod.
Let's calculate the change in length for each section separately:
1. Aluminum section:
The length of the aluminum section is given as 1.3m. The strain in this section (ε_al) can be calculated using Hooke's Law, which states that the strain is directly proportional to the applied stress:
ε_al = σ / E_al
where σ is the stress and E_al is the Young's modulus of aluminum.
We are given that the rod is under tension with a force of 5.7E10^3 N. To find the stress, we divide this force by the cross-sectional area of the rod. The radius is given as 0.2cm, so the diameter is 0.4cm.
Using the formula for the area of a circle (A = π*r^2), the cross-sectional area (A_al) can be calculated:
A_al = π * (0.2cm)^2
Next, calculate the stress (σ_al):
σ_al = F_al / A_al
where F_al is the applied force on the aluminum section.
Finally, substitute the values for stress and Young's modulus (E_al) to calculate the strain (ε_al). The Young's modulus for aluminum is approximately 7.0E10^10 N/m^2.
2. Copper section:
The length of the copper section is given as 2.6m. We can repeat the same calculations as in the aluminum section to find the strain in this section, using the appropriate Young's modulus for copper.
3. Total elongation:
Finally, to find the total elongation of the rod, we sum up the elongations in both sections:
Total elongation = elongation in aluminum section + elongation in copper section
Using the values calculated above, we can find the elongation of the rod.