A brass wire with Young's modulus of 9.20 1010 Pa is 2.5 m long and has a cross-sectional area of 4.6 mm2. If a weight of 5.2 kN is hung from the wire, by how much does it stretch?
E = 9 * 10^10
9 * 10^10 dL/2.5 = 5.2*10^3 /4.6*10^-6
9 dL /2.5 = .52/4.6
dL = .031m = 31mm = 3.1 cm
To find the amount of stretch in the wire, we can use Hooke's law, which states that the amount of stretch or compression of a material is directly proportional to the applied force, given the material's Young's modulus.
Hooke's law can be expressed as:
ΔL = (F * L) / (A * E)
Where:
ΔL = change in length (stretch or compression)
F = applied force
L = initial length of the wire
A = cross-sectional area of the wire
E = Young's modulus of the material
Given:
F = 5.2 kN (convert to N: 1 kN = 1000 N)
L = 2.5 m
A = 4.6 mm^2 (convert to m^2: 1 mm^2 = 10^-6 m^2)
E = 9.20 * 10^10 Pa
Let's calculate the amount of stretch:
ΔL = (F * L) / (A * E)
= (5200 N * 2.5 m) / (4.6 * 10^-6 m^2 * 9.20 * 10^10 Pa)
Solver the equation to get the value of ΔL:
= 3 * 10^-4 m
Therefore, the wire stretches by 0.0003 m or 0.3 mm when a weight of 5.2 kN is hung from it.
To calculate the amount by which the brass wire stretches when a weight is hung from it, we can use Hooke's Law. Hooke's Law states that the amount of deformation or stretch of an object is directly proportional to the force applied to it, provided the force does not exceed the elastic limit of the material.
The formula for Hooke's Law is:
ΔL = (F * L) / (A * Y)
Where:
ΔL is the change in length or the stretch of the wire
F is the force applied (in Newtons)
L is the original length of the wire
A is the cross-sectional area of the wire
Y is Young's modulus of the material
From the given data, we have:
F = 5.2 kN = 5.2 * 10^3 N (converting kilonewtons to newtons)
L = 2.5 m
A = 4.6 mm^2 = 4.6 * 10^-6 m^2 (converting square millimeters to square meters)
Y = 9.20 * 10^10 Pa
Substituting these values into the formula, we get:
ΔL = (5.2 * 10^3 N * 2.5 m) / (4.6 * 10^-6 m^2 * 9.20 * 10^10 Pa)
Simplifying the equation:
ΔL = (1.3 * 10^4 Nm) / (4.6 * 10^-6 m^3 * 9.20 * 10^10 Pa)
ΔL = 2.83 * 10^-5 m
Therefore, the brass wire stretches by approximately 2.83 * 10^-5 meters when a weight of 5.2 kN is hung from it.