What is the minimum diameter of a copper wire able to support 5 kg mass without exceeding elastic limit?
To determine the minimum diameter of a copper wire able to support a 5 kg mass without exceeding its elastic limit, we need to consider the mechanical properties of copper and how they relate to this scenario.
The elastic limit is the maximum amount of stress a material can withstand before it permanently deform or breaks. To calculate the stress, we need to know the force applied and the cross-sectional area of the wire.
The stress (σ) can be obtained using the formula:
σ = F / A
Where:
σ = Stress (Pa or N/m²)
F = Applied force (N)
A = Cross-sectional area (m²)
Let's assume that the elastic limit of copper is 207 MPa (megapascals). To convert this to pascals, we multiply by 10^6 (1 MPa = 10^6 Pa), giving us a value of 207 x 10^6 Pa.
Next, we need to determine the maximum stress the copper wire can withstand without exceeding the elastic limit. Rearranging the stress formula:
σ = F / A
A = F / σ
For a force of 5 kg in Earth's gravity, F = m x g, where g is the acceleration due to gravity (9.8 m/s²). So, F = 5 kg x 9.8 m/s².
Substituting these values into the area formula, we find:
A = (5 kg x 9.8 m/s²) / (207 x 10^6 Pa)
To find the minimum diameter of the wire, we need to calculate the cross-sectional area (A). For a round wire, the area can be found using the formula:
A = πr^2
Where:
A = Cross-sectional area (m²)
π = Pi (approximately 3.14)
r = Radius of the wire (m)
We can rearrange this formula to solve for the radius:
r = √(A / π)
Substituting the value of A calculated earlier, we have:
r = √(((5 kg x 9.8 m/s²) / (207 x 10^6 Pa)) / π)
Finally, to find the minimum diameter, we double the radius:
Minimum Diameter = 2 x r
By performing these calculations, you can determine the minimum diameter of the copper wire required to support a 5 kg mass without exceeding its elastic limit.