Calculating vector component problem: a heading of 140 degrees relative to North...

[wduk]

Hello

I am trying to calculate the components for a vector but am getting it wrong according to my book, though the book does not explain its reasoning for its calculation.

I have a heading of 140 degrees relative to North with a velocity of 1.1ms^1. So this was my calculation:

To get the acute angle of my triangle i did 180 - 140 = 40 degrees

Then using the equation of |v| = |v|cos(theta)i + |v|sin(theta)j

I got:
1.1cos(theta)i + 1.1sin(theta)j

Which gives: 0.8i + 0.7j, but since the heading is decreasing on the Y axis it becomes 0.8i - 0.7j

How ever my book seems to flip the functions around so it is doing:
1.1sin(theta)i + 1.1cos(theta)j

I don't understand why though? Cosine deals with the x position is what i always remember, and since i is used for east and j for north (y axis) why would the book flip the functions around in the calculation?

 

[Ishuda]

Hello

I am trying to calculate the components for a vector but am getting it wrong according to my book, though the book does not explain its reasoning for its calculation.

I have a heading of 140 degrees relative to North with a velocity of 1.1ms^1. So this was my calculation:

To get the acute angle of my triangle i did 180 - 140 = 40 degrees

Then using the equation of |v| = |v|cos(theta)i + |v|sin(theta)j

I got:
1.1cos(theta)i + 1.1sin(theta)j

Which gives: 0.8i + 0.7j, but since the heading is decreasing on the Y axis it becomes 0.8i - 0.7j

How ever my book seems to flip the functions around so it is doing:
1.1sin(theta)i + 1.1cos(theta)j

I don't understand why though? Cosine deals with the x position is what i always remember, and since i is used for east and j for north (y axis) why would the book flip the functions around in the calculation?

It is in the interpretation of 'relative to North'. If North is zero degrees and clockwise is positive then we would have x=|v| sin(140$\displaystyle ^\circ$-90$\displaystyle ^\circ$) = |v| sin(50$\displaystyle ^\circ$) and y = |v|cos(50$\displaystyle ^\circ$)

 

[wduk]

So i would have to imagine a right angled triangle and use the Y-axis as that triangle's positive x-axis ? Is that what you mean? If not im not visually understanding it =/

 

[wduk]

Have i made sense here or am i still misunderstanding ? =/ Not 100% sure if what i said works in all cases?

 

[Harry_the_cat]

So i would have to imagine a right angled triangle and use the Y-axis as that triangle's positive x-axis ? Is that what you mean? If not im not visually understanding it =/

Draw a set of x and y axes. North is along the positive y-axis.

140 degrees is measured clockwise from North (the y-axis).

Draw the vector in that position and 1.1 units long.

Now complete your triangle and mark in the relevant acute angle. Find the x-coordinate and the y-coordinate of the point at the end of the vector.

Drawing a diagram will always help!

 

[wduk]

Sorry for late reply.

I drew it but i am still confused by it.

A bearing from north at 140 gives a negative y and a positive x:



So since cosine deals with x and sine deals with y shouldn't the calculation be the cosine subtract thee sine ? Like 1.1cos(theta)i - 1.1sin(theta)j ?

I am confused how i should be understanding this when referring to north as the y axis here =/