A 4.0-m-long steel wire has a cross-sectional area of 0.050cm2. Its proportional limit has a value of 0.0016 times its Young's modulus. Its breaking stress has a value of 0.0065 times its Young's modulus. The wire is fastened at its upper end and hangs vertically.
A. How great a weight can be hung from the wire without exceeding the proportional limit? Take the Young's modulus for the rod to be Y=2.0×1011Pa.
B. How much will the wire stretch under this load?
thank you!
A) F = 0.0016YA, F= 1600N
B) Delta L = FL/AY, Delta L = 0.0064 m or 6.4 mm
C) F = 0.0065YA, F = 6500N
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Just did this problem and it's correct, I got'chu.
To solve this problem, we need to use Hooke's Law, which states that the stress (force per unit area) in a material is directly proportional to the strain (the change in length per unit length) within the elastic limit.
First, let's calculate the proportional limit stress and the breaking stress for the given wire.
The proportional limit stress (σ_proportional) can be found by multiplying the Young's modulus (Y) with the given factor (0.0016).
σ_proportional = Y × 0.0016
Substituting the given value for Young's modulus:
σ_proportional = 2.0 × 10^11 Pa × 0.0016
Next, let's calculate the breaking stress (σ_breaking) by multiplying the Young's modulus (Y) with the given factor (0.0065).
σ_breaking = Y × 0.0065
Substituting the given value for Young's modulus:
σ_breaking = 2.0 × 10^11 Pa × 0.0065
Now, let's move on to part A of the question:
A. How great a weight can be hung from the wire without exceeding the proportional limit?
To calculate the weight that can be hung without exceeding the proportional limit, we need to find the maximum stress that the wire can handle.
Stress is defined as the force (weight) divided by the cross-sectional area of the wire.
σ = F / A
Using this formula, we can rearrange it to find the maximum weight (F_limit) without exceeding the proportional limit.
F_limit = σ_proportional × A
Substituting the given value for the cross-sectional area:
F_limit = σ_proportional × 0.050 cm^2
Since the weight is given in terms of grams, we can use the conversion 1 N = 1000 g to convert it to Newtons.
Finally, to calculate the weight in grams:
Weight_limit = F_limit / 9.8 m/s^2
Now, let's move on to part B of the question:
B. How much will the wire stretch under this load?
To calculate the stretch, we can use Hooke's Law:
Stress (σ) = Young's modulus (Y) × Strain (ε)
Strain (ε) = Change in length (ΔL) / Original length (L)
Since the wire is hanging vertically, the change in length will be equal to the stretch, denoted as ΔL.
We can rearrange Hooke's Law to calculate the stretch:
ΔL = (σ / Y) × L
Substituting the calculated stress (σ) from part A:
ΔL = ( σ_proportional / Y) × L
Now we have the equation to calculate the stretch of the wire under the given load without exceeding the proportional limit.