1. Problem 1: A rope spans 34 feet between two solid supports, with a 296-pound weight attached at its midpoint causing the rope to dip 7 feet at the lowest point; compute the tension acting along the rope. 2. Consider one half of the rope to get the geometry: the horizontal half-span is 17 ft and the vertical drop is 7 ft. 3. Let $\theta$ be the angle the rope makes with the horizontal at the support; then $\tan(\theta)=\frac{7}{17}$. 4. Compute $\sin(\theta)$ by $\sin(\theta)=\frac{7}{\sqrt{7^2+17^2}}=\frac{7}{\sqrt{338}}$. 5. Vertical force balance for the weight gives $2T\sin(\theta)=296$ where $T$ is the tension magnitude along the rope. 6. Solve for $T$: $$T=\frac{296}{2\sin(\theta)}=\frac{296}{2\times\frac{7}{\sqrt{338}}}=\frac{296\sqrt{338}}{14}$$. 7. Numeric evaluation: $\sqrt{338}\approx18.38477631$ which gives $T\approx388.85$ lb. 8. Final answer for Problem 1: $T\approx388.85\ \text{lb}$. 9. Problem 2: A skier of mass 51.96 kg is pulled up a frictionless slope at angle 13.36° with the horizontal; given speed $v=2.01\ \text{m/s}$ and acceleration $a=0.268\ \text{m/s}^2$, find the magnitude $F_{rope}$ of the force from the rope. 10. Free-body along the slope: uphill tension $F_{rope}$ and downhill component of weight $mg\sin(\theta)$. 11. Newton's second law along the slope gives $F_{rope}-mg\sin(\theta)=ma$ so $F_{rope}=m\bigl(a+g\sin(\theta)\bigr)$. 12. Use $m=51.96\ \text{kg}$, $a=0.268\ \text{m/s}^2$, $\theta=13.36^\circ$, and take $g=9.8\ \text{m/s}^2$. 13. Compute $\sin(13.36^\circ)\approx0.23101$. 14. Evaluate $F_{rope}=51.96\times\bigl(0.268+9.8\times0.23101\bigr)\approx131.56$ N. 15. Final answer for Problem 2: $F_{rope}\approx131.56\ \text{N}$.