In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0kg from a vertical steel wire 0.50m long and 2.5 × 10^-3 cm^2 in cross sectional area. On the bottom of the sphere he attendes a similar steel wire, from which he hang a brass cube of mass 10.0kg. for each wire, compute (a) the tensile strain and (b) the elongation?
Start here:
young's modulus = stress/strain= (F/Area)/(changeL/Loriginal)
So look up youngs modulus for steel.
Then you know Force (equal to the total weight below), you know area, and you know Length original of that wire. Calculate changeL That gives you b) on both wires.
for strain, strain= (F/Area)/youngs modulus
To compute the tensile strain, we need to use the formula:
Strain = (change in length) / (original length)
(a) For the aluminum sphere:
The original length is the length of the wire, which is 0.50 m.
The change in length is the elongation.
We can find the elongation using Hooke's Law, which states:
Stress = Young's modulus × Strain
The stress on the wire can be calculated as the weight of the sphere:
Stress = mass × gravity
where mass = 6.0 kg and gravity = 9.8 m/s^2.
Now, we can find the strain:
Strain = Stress / Young's modulus
Young's modulus for aluminum is approximately 7 × 10^10 N/m^2.
Thus, we can calculate the tensile strain for the aluminum sphere.
(b) For the brass cube:
We'll follow the same procedure as above, but with different values.
The mass is 10.0 kg, and we'll need the Young's modulus for steel this time.
Now, we can calculate the tensile strain for the brass cube.
To find the elongation, we'll use the formula:
Elongation = Strain × original length
We'll calculate the elongation for both the aluminum sphere and the brass cube.
Let's plug in the values and calculate the results.
To compute the tensile strain and elongation for each wire, we'll use the following formulas:
(a) Tensile strain (ε) = (ΔL / L₀)
(b) Elongation (ΔL) = (F / (A × Y))
Where:
- ΔL is the change in length.
- L₀ is the original length.
- F is the force applied.
- A is the cross-sectional area.
- Y is the Young's modulus of the material.
Let's calculate the values step by step:
1. For the aluminum sphere:
- Mass (m) = 6.0 kg
- Cross-sectional area (A) = ?
- Young's modulus (Y) = ?
As the problem does not provide the cross-sectional area or Young's modulus of the aluminum sphere, we cannot calculate the tensile strain or elongation for this wire.
2. For the vertical steel wire:
- Mass (m) = ?
- Cross-sectional area (A) = 2.5 × 10^-3 cm^2 = 2.5 × 10^-7 m^2
- Young's modulus (Y) = ?
As the problem does not provide the mass or Young's modulus of the vertical steel wire, we cannot calculate the tensile strain or elongation for this wire either.
3. For the bottom steel wire:
- Mass (m) = 10.0 kg
- Cross-sectional area (A) = 2.5 × 10^-3 cm^2 = 2.5 × 10^-7 m^2
- Young's modulus (Y) = ?
Similar to the previous wires, the Young's modulus is not provided, so we need that information to calculate the tensile strain or elongation.
In summary, without the values for the Young's modulus of the materials, we cannot calculate the tensile strain or elongation for any of the wires.