- #1

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The solution manual shows the the vector resolved like

a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j

Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.

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- Thread starter th77
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- #1

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The solution manual shows the the vector resolved like

a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j

Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.

- #2

vanesch

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th77 said:to the Negative Y axis.

The solution manual shows the the vector resolved like

a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j

Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.

Because the angle is here defined differently than usual...

- #3

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[itex] \sin(x-90) = \cos(x) [/itex]

[itex] \cos(x-90) = \sin(x) [/itex]

- #4

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Thanks!! That leands me to another question...whozum said:

[itex] \sin(x-90) = \cos(x) [/itex]

[itex] \cos(x-90) = \sin(x) [/itex]

In this problem, the angle is 30 degrees with the negative axis so shouldn't cos 240 be equal to sin 30? They come to -0.5 and 0.5 respectively.

- #5

vanesch

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th77 said:Thanks!! That leands me to another question...

In this problem, the angle is 30 degrees with the negative axis so shouldn't cos 240 be equal to sin 30? They come to -0.5 and 0.5 respectively.

You're turning in the wrong direction: 30 degrees starting from the negative Y axis (=270 degrees) gives you 300 degrees, not 240...

- #6

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vanesch said:You're turning in the wrong direction: 30 degrees starting from the negative Y axis (=270 degrees) gives you 300 degrees, not 240...

the vector is in the 3rd quadrant so isn't it 240?

- #7

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