how to find the magnitudes of the x- and y-components of the tension?

Question

theeric · Answer

Does this relate to a specific problem?

anonymous · Answer

it says A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg. then asks for the x- and y- components of the tension. @theEric

theeric · Answer

Is there any other information in the problem?

unklerhaukus · Answer

use trigonometry and pythagorus

unklerhaukus · Answer

|dw:1389767913982:dw|

unklerhaukus · Answer

|dw:1389768038159:dw|

theeric · Answer

Should we assume the water to be a flat, frictionless surface?

anonymous · Answer

that all the info i have for it. 
& i think im supposed to assume no friction. @theEric

anonymous · Answer

& i meant that is asks for the magnitudes of the x- and y- components of the tension. @theEric

unklerhaukus · Answer

in my diagram the x- and y- components of the tension are labelled as T_x and T_y

theeric · Answer

Okay! Actually, the physical situation doesn't matter so much... You break up the tension like you break up forces!

theeric · Answer

Does that help?

anonymous · Answer

am i supposed to solve something in the diagram? @UnkleRhaukus

theeric · Answer

@UnkleRhaukus Provided a nice diagram.

anonymous · Answer

how would i find the magnitudes from the diagram? @theEric

unklerhaukus · Answer

|dw:1389768487498:dw|

anonymous · Answer

so y would be; 
490sin13 
& x is; 
490cos13 

and i just solve it like that? @UnkleRhaukus

theeric · Answer

Trigonometry! You know the angle. You'll have to use $\sin$ and $\cos$ to get the components, as @UnkleRhaukus illustrated in the diagrams.

unklerhaukus · Answer

yep, just make sure your calculator is set to degrees mode (not radians)

if you want to check the answers you get for the components , the squareroot sum of their squares should give you back T = 490 N

theeric · Answer

:)