If the elastic limit of steel is 5.0 multiplied by 108 Pa, determine the minimum diameter a steel wire can have if it is to support a 79 kg circus performer without its elastic limit being exceeded.
Maximum stress = (Force/Area)
5*10^8 N/m^2 = (79*9.8 N)/[(pi/4)D^2]
The mimimum diameter, D, will be in meters
pi/4 * D^2 = 1.55*10^-6 m^2
D^2 = 1.97*10^-6 m^2
D = 1.4*10^-3 m = 1.4 mm.
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Can you explain a little bit further
To solve this problem, we need to use the formula for stress, which is defined as the force applied divided by the cross-sectional area:
Stress = Force / Area
We can rearrange this formula to solve for the area:
Area = Force / Stress
Given:
Stress (elastic limit of steel) = 5.0 x 10^8 Pa
Force (weight of circus performer) = 79 kg x 9.8 m/s^2 (acceleration due to gravity) = 774.2 N
Substituting these values into the formula, we have:
Area = 774.2 N / (5.0 x 10^8 Pa)
Now, we need to use the formula for the area of a circle:
Area = π * r^2
Rearranging this formula to solve for the radius, we get:
r = sqrt(Area / π)
Substituting the previous expression for area into this formula, we have:
r = sqrt((774.2 N / (5.0 x 10^8 Pa)) / π)
Finally, to find the minimum diameter, we double the radius:
Diameter = 2 * r
Now we will calculate the minimum diameter:
Step 1: Calculate the area:
Area = 774.2 N / (5.0 x 10^8 Pa) = 1.5484 x 10^-6 m^2
Step 2: Calculate the radius:
r = sqrt((1.5484 x 10^-6 m^2) / π) ≈ 0.022 m
Step 3: Calculate the minimum diameter:
Diameter = 2 * r = 2 * 0.022 m = 0.044 m
Therefore, the minimum diameter the steel wire can have is approximately 0.044 meters to support the circus performer without exceeding its elastic limit.