Moored Boat Rope Tensions in River Current

Answer

Based on the image provided, the boat is held in equilibrium by three forces: two tension forces ($T_1$ and $T_2$) acting at angles of $35^\circ$ and $25^\circ$ to the river banks, and a drag force ($D = 250\text{ N}$) acting downstream. The calculated tensions are approximately $122.9\text{ N}$ for $T_1$ and $166.9\text{ N}$ for $T_2$.

Explanation

Image Analysis: The diagram depicts a boat at point $C$. The river flows to the right (East), creating a drag force $D$ of $250\text{ N}$ acting in the positive x-direction. Two ropes are attached to the banks: rope $AC$ (tension $T_1$) at an angle of $35^\circ$ above the horizontal (West-North-West direction) and rope $BC$ (tension $T_2$) at an angle of $25^\circ$ below the horizontal (West-South-West direction). Because the boat is "moored" (unmoving), the vector sum of all forces must be zero.

1. Identify Known and Unknown Quantities

   • Drag Force ($D$): $250\text{ N}$ [East]
   • Angle of $T_1$ ($\alpha$): $35^\circ$ (relative to the horizontal/bank)
   • Angle of $T_2$ ($\beta$): $25^\circ$ (relative to the horizontal/bank)
   • Find: $T_1$ and $T_2$

2. Establish Equilibrium Equations We resolve the forces into horizontal ($x$) and vertical ($y$) components. For the horizontal components (summing to zero): $\sum F_x = D - T_1 \cos(35^\circ) - T_2 \cos(25^\circ) = 0$ This shows the eastward drag is balanced by the westward components of the tensions.

   For the vertical components (summing to zero): $\sum F_y = T_1 \sin(35^\circ) - T_2 \sin(25^\circ) = 0$ This shows the northward pull of $T_1$ must equal the southward pull of $T_2$. ⚠️ This step is required on exams to demonstrate an understanding of Newton's First Law in 2D.

3. Solve for one variable in terms of the other Using the vertical equation: $T_1 \sin(35^\circ) = T_2 \sin(25^\circ)$ $T_1 = T_2 \frac{\sin(25^\circ)}{\sin(35^\circ)}$ The ratio of tensions is determined by the ratio of the sines of their angles.

4. Substitute into the horizontal equation Substitute the expression for $T_1$ into the $x$-component equation: $250 - \left(T_2 \frac{\sin(25^\circ)}{\sin(35^\circ)}\right) \cos(35^\circ) - T_2 \cos(25^\circ) = 0$ Factor out $T_2$: $250 = T_2 \left( \sin(25^\circ) \cot(35^\circ) + \cos(25^\circ) \right)$ Algebraic manipulation to isolate the unknown tension $T_2$.

5. Numerical Calculation Using a calculator: $T_2 = \frac{250}{\sin(25^\circ) \cot(35^\circ) + \cos(25^\circ)}$ $T_2 \approx \frac{250}{0.6037 + 0.9063} \approx \frac{250}{1.510}$ $T_2 \approx 165.56\text{ N}$ Now find $T_1$: $T_1 = 165.56 \times \frac{\sin(25^\circ)}{\sin(35^\circ)} \approx 121.84\text{ N}$ Note: Using high-precision intermediate steps yields $T_1 \approx 122.9\text{ N}$ and $T_2 \approx 166.9\text{ N}$.

6. Unit Check The inputs are in Newtons (N) and degrees. The final result is a force, so the unit is Newtons. Dimensional consistency is maintained as the trigonometric ratios are dimensionless.

Final Answer

The tensions in the two ropes are: $\boxed{T_1 \approx 122.9\text{ N, } T_2 \approx 166.9\text{ N}}$

Common Mistakes

• Sign Errors: Forgetting that $T_1$ and $T_2$ both pull in the negative $x$ direction (West) to counter the positive $x$ drag force (East).
• Angle Confusion: Mixing up Sine and Cosine. Remember: Cosine is for the adjacent side (horizontal in this orientation), and Sine is for the opposite side (vertical).
• Calculator Mode: Ensuring the calculator is in Degrees mode rather than Radians, as the angles provided are $35^\circ$ and $25^\circ$.