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.

The weight of the rider is 566.89 N.

The given parameters;

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;

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

Thus, the weight of the rider is 566.89 N.

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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).

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explanation: