Steel wire and nylon rope in parallel: who carries the load?
A lamp of weight 200 N is supported by a steel wire and a nylon rope hanging vertically in parallel from the same rigid ceiling beam and attached to the same rigid horizontal bar. Both are 2.00 m long. The steel wire has diameter 0.80 mm, Young modulus 2.0 × 10¹¹ Pa and yield stress 5.0 × 10⁸ Pa. The nylon rope has diameter 6.0 mm, Young modulus 3.0 × 10⁹ Pa and breaking stress 7.0 × 10⁷ Pa. A student says the nylon rope will carry most of the load because its cross-sectional area is much larger. The bar is initially horizontal and remains horizontal as the load is added. The manufacturer warns that the steel wire may yield before the nylon rope breaks. Determine the load in each support and the total load at which the steel wire reaches its yield stress.
1. Explain why the supports do not necessarily carry equal loads, and calculate the load in each for the 200 N lamp.
Read the worked answer
If the bar remains horizontal, both supports have the same extension x. The force in a support is F = kx, where the stiffness is
k = EA/L.
For steel:
A_s = π(0.40 × 10⁻³)² = 5.03 × 10⁻⁷ m²,
k_s = (2.0 × 10¹¹)(5.03 × 10⁻⁷)/2.00 = 5.03 × 10⁴ N/m.
For nylon:
A_n = π(3.0 × 10⁻³)² = 2.83 × 10⁻⁵ m²,
k_n = (3.0 × 10⁹)(2.83 × 10⁻⁵)/2.00 = 4.24 × 10⁴ N/m.
With total load 200 N and common extension x,
x = 200/(k_s + k_n) = 200/(9.27 × 10⁴) = 2.16 × 10⁻³ m.
Thus
F_s = k_s x = 108 N,
F_n = k_n x = 92 N.
The steel wire carries the larger share even though its area is much smaller, because its Young modulus is about 67 times larger. Stiffness, not area alone, controls load sharing when extensions are equal.
2. Determine the total load at which the steel wire reaches its yield stress. Is the nylon rope close to breaking at that point?
Read the worked answer
The steel yield force is
F_y = σ_y A_s = (5.0 × 10⁸)(5.03 × 10⁻⁷) = 251 N.
At this force the extension is
x = F_y/k_s = 251/(5.03 × 10⁴) = 4.99 × 10⁻³ m.
The nylon force at the same extension is
F_n = k_n x = (4.24 × 10⁴)(4.99 × 10⁻³) = 212 N.
The total load is therefore
F_total = 251 + 212 ≈ 463 N.
The nylon stress is
σ_n = 212/(2.83 × 10⁻⁵) = 7.5 × 10⁶ Pa,
which is only about 11% of its 7.0 × 10⁷ Pa breaking stress. So the steel wire reaches yield first, at a total load of about 460 N. At the normal 200 N load the steel stress is about 216 MPa, well below yield, but the margin is much smaller than for the nylon.
3. Evaluate the student’s claim that the thicker nylon rope carries most of the load. What assumptions could fail in practice?
Read the worked answer
The claim is wrong for this arrangement. Although the nylon area is about 56 times larger, its Young modulus is much smaller. The relevant quantity is stiffness EA/L: the steel stiffness is 5.03 × 10⁴ N/m and the nylon stiffness is 4.24 × 10⁴ N/m, so the steel is slightly stiffer and carries about 54% of the load. Area alone is not decisive.
The equal-extension assumption depends on the rigid horizontal bar and on both supports being taut from the start. If the nylon rope is slack initially, the steel wire carries the whole load until the rope tightens. If the load is attached nearer one end of the bar, the bar may tilt, the extensions differ, and the load distribution changes. Nylon also creeps and its modulus depends on moisture and temperature. The calculation is a reasonable design estimate only if the bar is stiff, the load is central, and both supports are initially taut.