Let's break down the problem step by step. The time-varying radial load is given by: \[ F = (100 - 100\cos{2\pi N}) \text{ kN} \] where \( N \) is the fraction of revolutions. We are asked to determine the **mean cubic load** \( F_e \). ## Step 1: Mean Cubic Load Formula The mean cubic load for a varying load \( F(N) \) over one revolution (from \( N = 0 \) to \( N = 1 \)) is given by: \[ F_e = \left( \int_0^1 [F(N)]^3 dN \right)^{1/3} \] ## Step 2: Substitute the Expression \[ F(N) = 100 - 100\cos{2\pi N} \] \[ F(N)^3 = [100 - 100\cos{2\pi N}]^3 \] Expand: \[ = 100^3 [1 - \cos{2\pi N}]^3 = 1,000,000 [1 - \cos{2\pi N}]^3 \] ## Step 3: Expand the Cubic Term \[ (1 - \cos{2\pi N})^3 = 1 - 3\cos{2\pi N} + 3\cos^2{2\pi N} - \cos^3{2\pi N} \] ## Step 4: Integrate Each Term Over \( N \) from 0 to 1 - \( \int_0^1 1 dN = 1 \) - \( \int_0^1 \cos{2\pi N} dN = 0 \) (over a full period) - \( \int_0^1 \cos^2{2\pi N} dN = \frac{1}{2} \) (average value over a period) - \( \int_0^1 \cos^3{2\pi N} dN = 0 \) (over a full period) So, \[ \int_0^1 (1 - 3\cos{2\pi N} + 3\cos^2{2\pi N} - \cos^3{2\pi N}) dN = 1 - 0 + 3 \cdot \frac{1}{2} - 0 = 1 + 1.5 = 2.5 \] ## Step 5: Plug Back into the Mean Cubic Load Formula \[ F_e = \left[ 1,000,000 \times (2.5) \right]^{1/3} = (2,500,000)^{1/3} \] ## Step 6: Calculate the Cube Root \[ 2,500,000^{1/3} = 135.7 \text{ kN} \] **Rounded to the nearest integer, the answer is:** \[ \boxed{136 \text{ kN}} \] --- ### **Final Answer:** **136 kN**