Let's analyze and solve the problem: **Given:** \[ F = (100 - 100 \cos 2\pi N) \text{ kN} \] where \( N \) is the fraction of revolutions. We are to find the **mean cubic load** \( F_m \). ### Step 1: Formula for Mean Cubic Load The mean cubic load for a time-varying load is given by: \[ F_m = \left[ \frac{1}{T} \int_0^T F^3 \, dN \right]^{1/3} \] Here, \( T \) is the period (for one complete revolution, \( N \) varies from 0 to 1). ### Step 2: Substitute the Given Load \[ F = 100 - 100 \cos(2\pi N) \] So, \[ F^3 = [100 - 100 \cos(2\pi N)]^3 \] ### Step 3: Compute the Integral Let's expand and integrate: \[ F_m = \left[ \int_0^1 [100 - 100 \cos(2\pi N)]^3 dN \right]^{1/3} \] Expanding the cube: \[ [100 - 100 \cos(2\pi N)]^3 = 100^3 [1 - \cos(2\pi N)]^3 \] \[ = 100^3 [1 - 3\cos(2\pi N) + 3\cos^2(2\pi N) - \cos^3(2\pi N)] \] Now integrate term by term from \( N = 0 \) to 1: - \(\int_0^1 1\,dN = 1\) - \(\int_0^1 \cos(2\pi N)\,dN = 0\) - \(\int_0^1 \cos^2(2\pi N)\,dN = \frac{1}{2}\) - \(\int_0^1 \cos^3(2\pi N)\,dN = 0\) (since \(\cos^3\) is odd-symmetric over one period) So, the sum is: \[ \int_0^1 [1 - 3\cos(2\pi N) + 3\cos^2(2\pi N) - \cos^3(2\pi N)] dN = 1 - 0 + 3 \cdot \frac{1}{2} - 0 = 1 + 1.5 = 2.5 \] ### Step 4: Plug In Values \[ F_m = \left[ 100^3 \cdot 2.5 \right]^{1/3} \] \[ = \left[ 1,000,000 \cdot 2.5 \right]^{1/3} \] \[ = \left[ 2,500,000 \right]^{1/3} \] \[ = 135.7 \approx 136 \text{ kN} \] ### **Final Answer:** \[ \boxed{136 \text{ kN}} \] So the answer is the first option: **136 kN**.