To determine the weight of the parasailor, we need to analyze the forces acting on the parasailor and apply some principles of vector mechanics. There are three main forces to consider:

1. The tension in the rope from the boat ($T_B = 1068 \, \text{N}$), which makes an angle of $12^\circ$ with the horizontal.
2. The tension in the rope from the parachute-like sail ($T_S$), which makes an angle of $34^\circ$ with the horizontal.
3. The weight of the parasailor ($W$), acting vertically downward due to gravity.

Since the parasailor is being towed at a constant speed, the forces in both the horizontal and vertical directions must balance. Therefore, we can break this problem down into horizontal ($x$) and vertical ($y$) components of the forces and use equilibrium conditions to find the weight of the parasailor.

Step 1: Break the forces into components

Horizontal components

• The horizontal component of the force from the boat: $T_B^x = T_B \cos(12^\circ)$
• The horizontal component of the force from the sail: $T_S^x = T_S \cos(34^\circ)$ Since the parasailor is moving at a constant speed, the net horizontal force is zero, so the horizontal forces must balance: $T_B^x = T_S^x$ This gives: $T_B \cos(12^\circ) = T_S \cos(34^\circ)$ Substitute $T_B = 1068 \, \text{N}$: $1068 \cos(12^\circ) = T_S \cos(34^\circ)$ Solving for $T_S$: $T_S = \frac{1068 \cos(12^\circ)}{\cos(34^\circ)}$ $T_S \approx \frac{1068 \times 0.9781}{0.8290} \approx 1259.4 \, \text{N}$

Step 2: Vertical force equilibrium

Now, in the vertical direction, the forces must also balance. The upward forces are the vertical components of the tension in the ropes from the boat and the sail, while the downward force is the weight of the parasailor.

The vertical components of the forces are:

• From the boat: $T_B^y = T_B \sin(12^\circ)$
• From the sail: $T_S^y = T_S \sin(34^\circ)$
• The weight of the parasailor acts downward: $W$

The net vertical force is zero, so the vertical forces must balance: $T_B^y + T_S^y = W$ Substituting the values for $T_B$ and $T_S$: $1068 \sin(12^\circ) + 1259.4 \sin(34^\circ) = W$ Calculate the vertical components: $1068 \sin(12^\circ) \approx 1068 \times 0.2079 = 221.9 \, \text{N}$ $1259.4 \sin(34^\circ) \approx 1259.4 \times 0.5592 = 703.9 \, \text{N}$ Now sum these components to find the weight: $W = 221.9 + 703.9 = 925.8 \, \text{N}$

Thus, the weight of the parasailor is approximately $925.8 \, \text{N}$.

Summary:

The weight of the parasailor is $W \approx 925.8 \, \text{N}$.

Would you like a breakdown of the steps or have any questions on the approach?

Here are some related questions you could explore:

1. How do you calculate the mass of the parasailor given the weight?
2. What happens if the angle of the boat rope increases while keeping the same tension?
3. How would the force balance change if the parasailor accelerates?
4. What is the importance of constant speed in this problem?
5. How would adding wind resistance affect the force balance?

Tip: When solving force equilibrium problems, breaking the forces into components along common directions (horizontal and vertical) simplifies the analysis.