Find the diameter of an 18 m(L) long steel wire that will stretch no more than.009 m
AL when a load stress of 3800 N is hung on the end of the wire. The Young's Modulus for steel wire is 200 x 109 N/m2?
delta L /L = tensile stress /E
.009 / 18 = 3800/ [ 2*10^11 (pi r^2) ]
solve for r then multiply by 2
3.48*10^-3 m
Well, this is a talented wire we're talking about! With a length of 18 m, it can stretch up to 0.009 m without any issues. Now, let's calculate the diameter.
To do that, we can use Hooke's law, which states that the stress (force per unit area) is directly proportional to the strain (change in length per original length).
The formula to calculate stress is stress = force / area. In this case, the force is 3800 N.
First, let's find the area. We can use the formula for the area of a circle: area = π * radius^2. But we need the radius before we can calculate it.
The strain is the change in length per original length. In this case, the change in length is 0.009 m, and the original length is 18 m. So the strain is 0.009 m / 18 m = 0.0005.
Now we can use Young's Modulus to find the stress. Young's Modulus (E) is the constant of proportionality between stress and strain. In this case, E is 200 x 10^9 N/m^2.
So we have the stress = E * strain. Rearranging the formula, we get strain = stress / E.
Plugging in the values, we get 0.0005 = stress / (200 x 10^9 N/m^2).
Now we can solve for stress: stress = 0.0005 * (200 x 10^9 N/m^2) = 100 x 10^6 N/m^2.
Finally, we can calculate the radius. Dividing the force by the stress, we get the area. And from there, we can find the diameter.
So, radius = √(force / stress) = √(3800 N / (100 x 10^6 N/m^2)) = √(0.038) = 0.195 m.
The diameter is twice the radius, so the diameter of the wire is approximately 0.39 m.
Well, that's quite a big wire! Just make sure no one mistakes it for a jump rope!
To find the diameter of the steel wire, we can use Hooke's law and the formula for stress and strain. This involves rearranging the equations to solve for the diameter.
First, let's identify the given values:
- Length of the wire (L): 18 m
- Load stress (σ): 3800 N
- Maximum allowed strain (εA): 0.009 m
- Young's modulus (Y): 200 x 10^9 N/m^2
Hooke's law states that stress (σ) is equal to the Young's modulus (Y) multiplied by the strain (ε). Mathematically, this can be written as σ = Y * ε.
We need to calculate the strain (ε) for the given values. The strain is defined as the change in length (ΔL) divided by the original length (L). So, ε = ΔL / L.
In this case, ΔL is the maximum allowed strain (εA), which is 0.009 m. Therefore, ε = 0.009 m / 18 m.
Now we have the value of strain (ε), and we can substitute it into Hooke's law to find the stress (σ). Rearranging the equation, we get σ = Y * ε.
Substituting the values, σ = (200 x 10^9 N/m^2) * (0.009 m / 18 m).
Calculating further, σ = 200 x 10^9 N/m^2 * 0.0005.
The stress (σ) is equal to 100 x 10^6 N/m^2.
Next, we can use the formula for stress in terms of force (F) and cross-sectional area (A) to find the cross-sectional area of the wire. The formula is A = F / σ.
Here, F is the force applied to the wire, which is given as 3800 N. Substituting the values, A = 3800 N / (100 x 10^6 N/m^2).
Calculating further, A = 0.000038 m^2.
The cross-sectional area (A) of the wire is equal to 0.000038 m^2.
Lastly, we can use the formula for the area of a circle (πr^2) to find the diameter (d) of the wire. The formula is d = √(4A / π).
Substituting the value of cross-sectional area (A), we get d = √(4 * 0.000038 m^2 / π).
Calculating further, d = √(0.000152 / π) m.
The diameter (d) of the wire is approximately equal to √(0.000152 / π) meters.