Considering forces in horizontal direction,
F cos(30) = T cos(17)
F = 1700 cos(17)/cos(30)
F = 1877.22 N
Now taking vertical direction,
mg = (F/2) - T sin(theta)
Weight, W = (1877.22/2) - (1700 x sin(17))
W = 441.58 N
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there is no picture PHY <CH 05 HW Problem 5.2 In the sport of parasailing, a...
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 17 ∘ from horizontal. The tension in the rope is 1500 N. The force of the sail on the rider is 30∘ from horizontal.
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 degrees from horizontal. The tension in the rope is 1900 NN. The force of the sail on the rider is 30 degrees from horizontal. What is the weight of the rider?
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. Include a...