1. **Problem statement:** A block is held on a smooth inclined plane by a string pulling parallel to the plane with a force of 27.3 pounds. The plane is inclined at an angle of 24° 50'. We need to find the weight $W$ of the block. 2. **Understanding the forces:** Since the plane is smooth (frictionless), the only forces along the plane are the component of the block's weight down the plane and the tension $T$ in the string pulling up the plane. 3. **Formula:** The component of the weight along the plane is $W \sin \theta$, where $\theta = 24^\circ 50'$. Since the block is held stationary, the tension balances this component: $$T = W \sin \theta$$ 4. **Calculate $\sin 24^\circ 50'$:** Convert $50'$ to decimal degrees: $$50' = \frac{50}{60} = 0.8333^\circ$$ So, $$\theta = 24 + 0.8333 = 24.8333^\circ$$ Using a calculator, $$\sin 24.8333^\circ \approx 0.419$$ 5. **Solve for $W$:** $$W = \frac{T}{\sin \theta} = \frac{27.3}{0.419}$$ 6. **Intermediate step showing cancellation:** $$W = \frac{27.3}{\cancel{0.419}} \times \frac{1}{\cancel{0.419}}$$ 7. **Final calculation:** $$W \approx 65.2$$ 8. **Answer:** The weight of the block is approximately 65.2 pounds.