Let's break down the problem: Given: \[ F = (100 - 100\cos{2\pi N}) \text{ kN} \] where \( N \) is the fraction of revolutions. **Task**: Determine the **mean cubic load** \( F_m \). --- ### Step 1: Mean Cubic Load Formula The mean cubic load for a time-varying load is given by: \[ F_m = \sqrt[3]{\frac{1}{T} \int_0^T F(t)^3 dt} \] Since the load is a function of \( N \), and \( N \) varies from 0 to 1 for one full revolution: \[ F_m = \sqrt[3]{\int_0^1 F(N)^3 dN} \] --- ### Step 2: Substitute the Given Expression \[ F(N) = 100 - 100\cos(2\pi N) \] \[ F(N)^3 = (100 - 100\cos(2\pi N))^3 \] --- ### Step 3: Expand the Cubic Expression Let \( x = \cos(2\pi N) \), then: \[ (100 - 100x)^3 = 100^3 (1 - x)^3 = 10^6(1 - 3x + 3x^2 - x^3) \] --- ### Step 4: Integrate Over One Revolution \[ F_m^3 = \int_0^1 10^6 (1 - 3x + 3x^2 - x^3) dN \] But \( x = \cos(2\pi N) \). So: \[ F_m^3 = 10^6 \int_0^1 [1 - 3\cos(2\pi N) + 3\cos^2(2\pi N) - \cos^3(2\pi N)] dN \] Integrate each term separately: --- #### Integrals: - \(\int_0^1 1\,dN = 1\) - \(\int_0^1 \cos(2\pi N)\,dN = 0\) (over full period) - \(\int_0^1 \cos^2(2\pi N)\,dN = \frac{1}{2}\) - \(\int_0^1 \cos^3(2\pi N)\,dN = 0\) (over full period) --- ### Step 5: Substitute \[ F_m^3 = 10^6 [1 - 0 + 3 \times \frac{1}{2} - 0] \] \[ F_m^3 = 10^6 [1 + 1.5] = 10^6 \times 2.5 = 2.5 \times 10^6 \] --- ### Step 6: Take the Cube Root \[ F_m = (2.5 \times 10^6)^{1/3} \] Calculate: \[ 2.5^{1/3} \approx 1.357 \] \[ (10^6)^{1/3} = 100 \] \[ F_m \approx 1.357 \times 100 = 135.7 \text{ kN} \] --- ### Step 7: Choose Closest Option **Closest option is:** \[ \boxed{136 \text{ kN}} \] --- ### **Final Answer** \[ \boxed{136 \text{ kN}} \]