A 1.1 x 10-3-kg house spider is hanging vertically by a thread that has a Young's modulus of 4.5 x 109 N/m2 and a radius of 16 x 10-6 m. Suppose that a 81-kg person is hanging vertically on an aluminum (Young's modulus 6.9 x 1010 N/m2) wire. What is the radius of the wire that would exhibit the same strain as the spider's thread, when the thread is stressed by the full weight of the spider?
Thanks in advance..
To find the radius of the wire that would exhibit the same strain as the spider's thread, we first need to calculate the strain in the spider's thread.
The strain (ε) is defined as the ratio of the change in length (ΔL) to the original length (L), or ε = ΔL / L.
In this case, the spider's thread is hanging vertically, so it is under tension due to the gravitational force. The stress (σ) in the spider's thread can be calculated using the formula σ = F / A, where F is the force and A is the cross-sectional area.
The force (F) exerted on the spider's thread can be calculated using the formula F = mg, where m is the mass of the spider and g is the acceleration due to gravity.
Given:
Mass of spider (m) = 1.1 x 10^(-3) kg
Acceleration due to gravity (g) = 9.8 m/s^2
Young's modulus of spider's thread (E) = 4.5 x 10^9 N/m^2
Radius of the spider's thread (r) = 16 x 10^(-6) m
Now, let's calculate the force exerted on the spider's thread:
F = mg
F = (1.1 x 10^(-3) kg) * (9.8 m/s^2)
F ≈ 1.078 x 10^(-2) N
Next, we need to calculate the cross-sectional area of the spider's thread:
The cross-sectional area of a thread is given by the formula A = πr^2, where r is the radius of the thread. Substituting the given radius, we get:
A = π * (16 x 10^(-6) m)^2
A ≈ 8.042 x 10^(-10) m^2
Now, we can calculate the stress (σ) in the spider's thread:
σ = F / A
σ = (1.078 x 10^(-2) N) / (8.042 x 10^(-10) m^2)
σ ≈ 1.341 x 10^(7) N/m^2
Now, we can use Hooke's Law to find the strain (ε) in the spider's thread. Hooke's Law states that the strain is directly proportional to the stress and inversely proportional to the Young's modulus. The formula is ε = σ / E, where σ is the stress and E is the Young's modulus.
Substituting the given values:
ε = (1.341 x 10^(7) N/m^2) / (4.5 x 10^9 N/m^2)
ε ≈ 2.980 x 10^(-3)
Now, let's find the radius of the wire that would exhibit the same strain as the spider's thread.
Using the same formula, we can rearrange it to solve for the radius (r):
ε = σ / E
σ = ε * E
F / A = ε * E
r = √(F / (A * ε * E))
Given:
Mass of person (m) = 81 kg
Young's modulus of aluminum wire (E) = 6.9 x 10^10 N/m^2
First, let's calculate the force exerted on the aluminum wire:
F = mg
F = (81 kg) * (9.8 m/s^2)
F ≈ 794.4 N
Next, we need to calculate the cross-sectional area of the aluminum wire. We don't know the radius, so we'll use the variable r here:
A = πr^2
Substituting the formula into the equation:
r = √(F / (A * ε * E))
r = √((794.4 N) / (π * r^2 * (2.980 x 10^(-3)) * (6.9 x 10^10 N/m^2)))
Rearranging the equation:
r = √(794.4 N / (π * (2.980 x 10^(-3)) * (6.9 x 10^10 N/m^2)))
r = √(794.4 N / (2.738 x 10^8 N/m^2))
r ≈ 1.92 x 10^(-2) m
Therefore, the radius of the wire that would exhibit the same strain as the spider's thread, when the thread is stressed by the full weight of the spider, is approximately 1.92 x 10^(-2) m.