a copper wire of length 2 metre and diameter 3 mm has its one end tied to a rigid support with the free end carrying a mass of 4Kg find the extension in the wire if the elastic limit of copper is 1.5*10^8 N/m^2 and it is not to be exceeded what should be the minimum diameter of the wire ? y for copper1.1*10^11
To find the extension in the wire, we can use Hooke's law, which states that the extension of a material is directly proportional to the applied force:
F = k * ΔL
Where F is the force applied, k is the spring constant, and ΔL is the extension in the wire.
Given that the mass of the object is 4 kg and the acceleration due to gravity is 9.8 m/s^2, the force applied is:
F = m * g
F = 4 kg * 9.8 m/s^2
F = 39.2 N
We can now find the spring constant using Young's modulus (y) and the cross-sectional area (A) of the wire:
k = (y * A) / L
Where L is the length of the wire.
The cross-sectional area can be found using the formula:
A = π * (d/2)^2
Where d is the diameter of the wire.
Substituting the given values:
A = π * (0.003 m / 2)^2
A = π * (0.0015 m)^2
A = 0.007065 m^2
Substituting this value and other given values into the equation for the spring constant:
k = (1.1 * 10^11 N/m^2 * 0.007065 m^2) / 2 m
k = 3.878 * 10^8 N/m
Now we can find the extension in the wire using Hooke's law:
ΔL = F / k
ΔL = 39.2 N / (3.878 * 10^8 N/m)
ΔL = 1.01 * 10^-7 m
Now let's find the minimum diameter of the wire that would prevent the elastic limit from being exceeded. The maximum tension force that the wire can withstand without exceeding the elastic limit is:
F_max = y * A
Substituting the given values:
F_max = (1.1 * 10^11 N/m^2) * (π * (d/2)^2)
F_max = 1.1 * 10^11 N/m^2 * π * (d/2)^2
Solving for d:
d/2 = sqrt(F_max / (1.1 * 10^11 N/m^2 * π))
d/2 = sqrt(39.2 N / (1.1 * 10^11 N/m^2 * π))
d/2 = 1.26 * 10^-4 m
Multiplying by 2 to get the full diameter:
d = 2 * 1.26 * 10^-4 m
d = 2.52 * 10^-4 m
Hence, the minimum diameter of the wire should be 2.52 * 10^-4 meters to prevent exceeding the elastic limit.