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A water skier is pulled behind a boat by a rope. The rope has a tension of 650 N and is at an angle of 27°. What is the y-component of the tension?

2 Answers

1 vote

Answer: 295N

Step-by-step explanation:

User Curline
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8.1k points
6 votes
There is no figure, so I asume the angle of the rope is measured with respect to the horizontal direction.

If this is the case, then the magnitude of the tension is the hypothenuse of a right triangle, of which
T_x and
T_y (horizontal and vertical component of the tension) are the other sides, and
\alpha is the angle between T and
T_x.
Therefore,
T_y (the side of the triangle opposite to
\alpha) is given by the magnitude of the tension multiplied by the sine of the angle:

T_y = T \sin \alpha = (650 N)(\sin 27^(\circ) )=295 N
User Sandinmyjoints
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8.1k points