Question

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?

Answer 1

Answer: 295N

Step-by-step explanation:

Answer 2

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`