1. **State the problem:** We have a barbell with two balls of mass $m=0.4$ kg at the ends of a rod of length $d=0.28$ m. Two forces of equal magnitude $F$ act on the balls in opposite directions, causing a change in angular momentum $\Delta L = 4.8 \times 10^{-4}$ kg·m$^2$/s over $\Delta t = 0.008$ s. The angular velocity is $114$ rad/s and the angle $\theta = 29^\circ$. We need to find the magnitude of each force $F$. 2. **Relevant formula:** Torque $\tau$ relates to the change in angular momentum by $$\tau = \frac{\Delta L}{\Delta t}$$ Torque is also given by $$\tau = F r \sin\theta$$ where $r$ is the lever arm (distance from axis to force application point), here $r = \frac{d}{2}$. 3. **Calculate torque:** $$\tau = \frac{4.8 \times 10^{-4}}{0.008} = 0.06 \text{ N·m}$$ 4. **Calculate lever arm:** $$r = \frac{d}{2} = \frac{0.28}{2} = 0.14 \text{ m}$$ 5. **Solve for force $F$:** $$\tau = F r \sin\theta \implies F = \frac{\tau}{r \sin\theta}$$ Substitute values: $$F = \frac{0.06}{0.14 \times \sin 29^\circ}$$ Calculate $\sin 29^\circ$: $$\sin 29^\circ \approx 0.4848$$ 6. **Intermediate step with cancellation:** $$F = \frac{0.06}{0.14 \times 0.4848} = \frac{0.06}{\cancel{0.14} \times 0.4848}$$ 7. **Final calculation:** $$F = \frac{0.06}{0.067872} \approx 0.883 \text{ N}$$ **Answer:** Each force has magnitude approximately $0.88$ N.