What hanging mass will stretch a 1.9-m-long, 0.54mm - diameter steel wire by 1.2mm ?
To find the hanging mass that will stretch the steel wire by 1.2 mm, we can use Hooke's Law, which states that the extension of an elastic material is directly proportional to the applied force. The formula for Hooke's Law is:
F = k * x
Where:
F is the applied force (weight of the hanging mass)
k is the spring constant (stiffness) of the wire
x is the extension of the wire
In this case, we need to find the hanging mass, so we can rearrange the formula as follows:
m = F / g
Where:
m is the mass of the hanging object
F is the applied force (weight of the hanging mass)
g is the acceleration due to gravity (approximately 9.8 m/s²)
To find the spring constant, we can use the formula for the spring constant of a wire:
k = (π * d^2 * E) / (4 * L)
Where:
d is the diameter of the wire
E is the Young's modulus of the material (for steel, it is around 200 GPa)
L is the original length of the wire
Now let's plug in the given values:
Original length of the wire (L) = 1.9 m
Diameter of the wire (d) = 0.54 mm = 0.54 * 10^-3 m
Extension of the wire (x) = 1.2 mm = 1.2 * 10^-3 m
Young's modulus of steel (E) = 200 GPa = 200 * 10^9 N/m²
First, let's calculate the spring constant (k):
k = (π * (0.54 * 10^-3)^2 * (200 * 10^9)) / (4 * 1.9)
k ≈ 3.83 * 10^3 N/m
Now let's calculate the applied force (F):
F = k * x
F = (3.83 * 10^3) * (1.2 * 10^-3)
F ≈ 4.60 N
Finally, let's calculate the hanging mass (m):
m = F / g
m = 4.60 / 9.8
m ≈ 0.47 kg
Therefore, a hanging mass of approximately 0.47 kg will stretch the 1.9-m-long, 0.54 mm-diameter steel wire by 1.2 mm.