Answer:
1.18 m
Explanation:
This problem can be divided into three parts: using stress and strain to find the ratio of the tension forces in each wire; balancing the forces on the beam to solve for the tension forces in each wire; and finally, balancing the moments (or torques) on the beam to solve for the position of the lamp.
The sum of the moments about a pivot point is equal to the moment of inertia times the angular acceleration. For a static body (no acceleration), the sum of moments about any point is zero. Therefore, we can pick a point (for example, point A), calculate the moment of each force about this point (moment = force × distance), and set the sum equal to 0. For this problem, there are four forces on the beam:
- Fs, the tension force in the steel wire
- Fa, the tension force in the aluminum wire
- mg, the weight of the beam
- Mg, the weight of the lamp
Given that the beam is horizontal (meaning the wires have equal deflection), we can use the formulas for stress and strain to solve for the tension force in each wire, then sum the moments about point A to find the position of the lamp.
The stress in each wire is equal to the force divided by the cross sectional area (σ = F/A). The stress is also equal to the Young modulus times the strain (σ = Eε), where strain is the deflection divided by the initial length (ε = ΔL/L₀). Substituting:
σ = σ
F/A = Eε
F/A = EΔL/L₀
F/(EA) = ΔL/L₀
Since both wires have the same deflection and the same initial length, then:
Fs / (Es As) = Fa / (Ea Aa)
Fs / (Es (¼ π Ds²)) = Fa / (Ea (¼ π Da²))
Fs / (Es Ds²) = Fa / (Ea Da²)
Plug in values:
Fs / ((210×10⁹ Pa) (0.80 mm)²) = Fa / ((70×10⁹ Pa) (1.60 mm)²)
Fs = 0.75 Fa
Applying Newton's second law of motion, the net force in the y direction is equal to the mass times acceleration (which is 0 for a static body).
∑F = ma
Fs + Fa − Mg − mg = 0
Fs + Fa = Mg + mg
0.75 Fa + Fa = (M + m) g
1.75 Fa = (M + m) g
Substituting:
1.75 Fa = (16.0 kg + 4.4 kg) (9.8 m/s²)
Fa = 114 N
Fs = 0.75 Fa = 86 N
Finally, we sum the moments about point A. Taking clockwise to be positive, and counterclockwise to be negative, the sum of moments is:
∑τ = Iα
(Fs) (0 m) + (mg) (1.00 m) + (Mg) (x) − (Fa) (2.00 m) = 0
Notice that the tension force Fs does not create a moment at point A, because it passes through that point, so the distance is 0 m. Also notice that the tension force Fa creates a counterclockwise moment, whereas the other forces create a clockwise moment.
Solving:
(mg) (1.00 m) + (Mg) (x) = (Fa) (2.00 m)
(4.4 kg × 9.8 m/s²) (1.00 m) + (16.0 kg × 9.8 m/s²) (x) = (114 N) (2.00 m)
43.12 Nm + (156.8 N) x = 228 Nm
(156.8 N) x = 184.88 Nm
x = 1.18 m
