A .5 m long, 3 cm diameter metal bar has a Young’s Modulus of 7.0x1010 N/m2. If a 50,000 N force is applied to the end of the bar, by how much will it compress?
To find the compression of the metal bar, we can use Hooke's Law, which states that the elongation or compression of an object is directly proportional to the applied force, as long as the limit of proportionality is not exceeded. The formula for Hooke's Law is:
ΔL = (F * L) / (A * E)
where:
ΔL is the change in length (compression or elongation),
F is the applied force,
L is the original length of the bar,
A is the cross-sectional area of the bar,
E is the Young's Modulus.
First, let's convert the diameter of the bar to radius.
Radius = diameter / 2 = 3 cm / 2 = 1.5 cm = 0.015 m
Using the formula for the area of a circle:
A = π * r^2
A = π * (0.015 m)^2 ≈ 0.0007068 m^2
Now we can substitute the values into the formula:
ΔL = (F * L) / (A * E)
ΔL = (50,000 N * 0.5 m) / (0.0007068 m^2 * 7.0x10^10 N/m^2)
ΔL ≈ 0.106 m
Therefore, the metal bar will compress by approximately 0.106 meters when a 50,000 N force is applied to its end.