A 148 kg student sits on a chair which is solely supported by a solid 0.97 meter-long steel rod 0.68 cm in diameter. What is the change in length of the rod produced by the student's weight?
Delta l =
To calculate the change in length of the rod, we need to use the formula for linear strain, which is given by:
Δl = (F * L) / (A * E)
Where:
Δl = Change in length of the rod
F = Force applied (weight of the student)
L = Original length of the rod
A = Cross-sectional area of the rod
E = Young's modulus of the material (in this case, steel)
First, let's find the cross-sectional area of the rod:
The diameter of the rod is given as 0.68 cm, so the radius is half of that, which is 0.34 cm. However, we need to convert it to meters, so it becomes 0.0034 meters.
The formula for the area of a circle is given by A = π * r², where r is the radius. So, plugging in the values:
A = π * (0.0034 m)² ≈ 0.00363 m²
The Young's modulus of steel can vary depending on the type of steel, but for general structural steel, it is around 200 GPa, which is equivalent to 200 x 10^9 Pa.
Now, let's calculate the change in length:
Δl = (F * L) / (A * E)
The weight of the student is given as 148 kg, so we need to multiply it by the acceleration due to gravity, which is approximately 9.8 m/s²:
F = 148 kg * 9.8 m/s² ≈ 1449.6 N
The original length of the rod is given as 0.97 meters.
Substituting all the values into the equation:
Δl = (1449.6 N * 0.97 m) / (0.00363 m² * 200 x 10^9 Pa)
Calculating the expression:
Δl ≈ 0.052 meters
So, the change in length of the rod produced by the student's weight is approximately 0.052 meters.