A 1.2m -long steel rod with a diameter of 0.50cm hangs vertically from the ceiling. An auto engine weighing 4.7kN is hung from the rod. By how much does the rod stretch?
so in my head I am supposed to have stored the Young's modulus of steel ?
delta L/L = tensile stress/E
or delta L = F L /(E A)
F = 4700 N
A = (pi/4) .005^2 meters^2
L = 1.2 meters
E = you look it up
You look it up.
To determine the amount by which the rod stretches, we need to consider the stress-strain relationship for the steel rod and apply Hooke's Law.
1. First, let's calculate the cross-sectional area of the rod, which is required for the stress calculation. The diameter of the rod is given as 0.50 cm. To find the area, we use the formula for the cross-sectional area of a circle: A = π * r^2, where r is the radius.
Given the diameter (d) as 0.50 cm, we can find the radius (r) by dividing the diameter by 2:
r = d/2 = 0.50 cm / 2 = 0.25 cm = 0.0025 m
Then, we can calculate the cross-sectional area (A) using the formula:
A = π * r^2 = π * (0.0025 m)^2 = 1.962 x 10^(-5) m^2
2. Next, we need to calculate the stress (σ) in the rod using the formula: stress = force / area. Given that the weight of the engine is 4.7 kN, we need to convert it to Newtons (N).
1 kN = 1000 N, so 4.7 kN = 4.7 * 1000 N = 4700 N
Now we can calculate the stress:
Stress (σ) = Force / Area = 4700 N / 1.962 x 10^(-5) m^2 = 2.395 x 10^8 N/m^2 (or 2.395 GPa)
3. Next, we need to determine the Young's modulus (E) of the steel rod. Young's modulus is a measure of the stiffness or elasticity of a material. For steel, the Young's modulus is typically around 200 GPa (Gigapascals), which is equivalent to 2 x 10^11 N/m^2.
4. Now we can use Hooke's Law, which states that the stress in a material is directly proportional to its strain (proportional limit assumed). We can rearrange the equation to solve for strain (ε), which represents the elongation or stretch in the rod.
Strain (ε) = Stress (σ) / Young's modulus (E)
Plugging in the values:
Strain (ε) = 2.395 GPa / 200 GPa = 0.011975
5. Finally, we can calculate the elongation or stretch (ΔL) of the rod using the equation: ΔL = ε * original length of the rod.
The original length of the rod is given as 1.2 m, so
ΔL = 0.011975 * 1.2 m = 0.01437 m (or 1.437 cm)
Therefore, the steel rod will stretch by approximately 0.01437 meters (or 1.437 centimeters) when the 4.7 kN engine is hung from it.