Question

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A rider is towed at a constant speed by a rope that is at an angle of 15 ∘ from horizontal. The tension in the rope is 1900 N. The force of the sail on the rider is 30∘ from horizontal. What is the weight of the rider? Express your answer with the appropriate units.

Answer 1

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

Answer 2

The weight of the rider is 566.89 N.

The given parameters;

• Tension in the rope, T = 1900 N

• angle of inclination of Tension, = 15⁰

• The force of sail on the rider, = F

• angle of inclination of Force, = 30 ⁰

Let the weight of the rider = W

Apply Newton's second law of motion, to determine the net force in horizontal and vertical direction.

F = ma

Sum of the forces in horizontal direction is calculated as follows;

$\Sigma F_x = 0\\\\Fcos(30) - Tcos(15) = 0\\\\0.866 F - 0.965T = 0\\\\0.866F = 0.965T\\\\F = \frac{0.965T}{0.866} , \ \ T = 1900 \ N\\\\F = \frac{0.965 \times 1900}{0.866} \\\\F = 2,177.21 \ N$

Sum of the forces in the vertical direction is calculated as follows;

$\Sigma F_y = 0\\\\Fsin(30) - W -Tsin(15) = 0\\\\W = Fsin(30)- Tsin(15)\\\\W = (2117.21 \times 0.5) - (1900\times 0.2588)\\\\W = 566.89 \ N$

Thus, the weight of the rider is 566.89 N.

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