A 215-kg load is hung on a wire of length of 3.50 m, cross-sectional area 2.000x10^-5 m2, and Young's modulus 8.00x 10^10 N/m2. What is its increase in length?
delta L/L = (T/area) / E
= (215*9.81 /2*10^-5) / 8*10^10
= ( 1054 * 10^5 ) / 8*10^10
= 132 * 10-5
so
delta L = (3.5)(132*10^-5)
= 461 * 10^-5
= .00461 meters = .461 cm
Well, let's see if we can stretch our imagination a little here (pun intended). To calculate the increase in length, we need to consider Hooke's law, which states that the change in length is directly proportional to the force applied.
Now, in this case, we have a load of 215 kg hung on a wire. But let's not get too caught up in details - we don't want the load to take us down a slippery slope! The force applied by the load can be calculated using the formula F = mg, where m is the mass and g is the acceleration due to gravity.
So, F = 215 kg × 9.8 m/s² = 2107 N. Now that we've got our force, we can calculate the increase in length using Hooke's law, which states that the force applied is equal to the product of the spring constant (which is Young's modulus) and the change in length divided by the original length.
So, F = (Y × A × ΔL) / L, where Y is Young's modulus, A is the cross-sectional area, ΔL is the change in length, and L is the original length.
Rearranging this equation, we can solve for ΔL: ΔL = (F × L) / (Y × A).
Plugging in the values: ΔL = (2107 N × 3.5 m) / (8.00 × 10^10 N/m² × 2.00 × 10^-5 m²).
Doing some calculations (or letting a calculator do it for us), we find that the increase in length is approximately 0.0001535 meters... or about the length of a eyelash, so don't blink or you might miss it!
To find the increase in length of the wire, we can use Hooke's Law, which states that the extension of an elastic object is directly proportional to the load applied to it. The formula for Hooke's Law is:
F = kΔL
Where:
F = force applied (load)
k = spring constant (Stress/Strain)
ΔL = change in length
To find the spring constant (k), we can use Young's modulus (Y) and the formula:
k = Y * (A / L)
Where:
Y = Young's modulus
A = cross-sectional area
L = original length
Let's calculate the spring constant (k) first:
k = (8.00 * 10^10 N/m^2) * (2.000 * 10^-5 m^2 / 3.50 m)
k ≈ 4.571 * 10^5 N/m
Now, we can find the change in length (ΔL) using Hooke's Law:
ΔL = F / k
F = 215 kg * 9.8 m/s^2 (weight)
ΔL = (215 kg * 9.8 m/s^2) / (4.571 * 10^5 N/m)
ΔL ≈ 0.464 m (rounded to three decimal places)
Therefore, the increase in length of the wire is approximately 0.464 meters.
To find the increase in length of the wire when the load is hung, we can use Hooke's Law, which states that the extension or deformation of a linearly elastic material is directly proportional to the force applied to it.
Hooke's Law equation:
ΔL = (F * L) / (A * E)
Where:
ΔL = Change in length (increase) of the wire
F = Applied force (weight of the load)
L = Original length of the wire
A = Cross-sectional area of the wire
E = Young's modulus of the wire
Given values:
F = 215 kg (weight of the load) * 9.8 m/s^2 (acceleration due to gravity) = 2107 N
L = 3.50 m
A = 2.000 x 10^-5 m^2
E = 8.00 x 10^10 N/m^2
Now we can substitute the values into the equation:
ΔL = (2107 N * 3.50 m) / (2.000 x 10^-5 m^2 * 8.00 x 10^10 N/m^2)
ΔL = 0.01556125 m (rounded to 5 decimal places)
Therefore, the increase in length of the wire when the load is hung is approximately 0.0156 meters.