Translating your directions into the standard trig notation, where East is 0°
and counterclockwise would be a positive rotation ....
15° south of west --- to me meant: 195°
12° east of south--- to me meant : 282°
so (3cos195,3sin195) + (4cos282,4sin282)
= (-2.066... , -4.689..)
I won't even attempt to figure out where you got your angles of 255 and 168 from.
You must have taken North to be 0° and rotate clockwise?
Question: Add the vectors using components. 3 N in a direction 15° south of west and 4 N in a direction 12° east of south.
I have attempted the question and got a resultant of (3.136, -2.06) from (3cos255°, 3sin255°)+(-4cos168°,4sin168°). I do not know if I did it correctly. Please provide a step-by-step solution for me to understand how to do these type of questions on my own. Thanks!
3 answers
If you draw the diagram, you can see that your first vector 3@155° puts you in the third quadrant at (-3cos15°,-3sin15°)
Measuring u° clockwise (from due North = 0°), the standard angle of v° (counterclockwise from the +x direction) is v = 90°-u. That would give you
(3cos(90-255)°,3sin(90-255)°) = (3cos(-165°),3sin(-165°)) = (-3cos15°,-3sin15°)
So you have the right idea, but you need to translate between the two orientations.
Measuring u° clockwise (from due North = 0°), the standard angle of v° (counterclockwise from the +x direction) is v = 90°-u. That would give you
(3cos(90-255)°,3sin(90-255)°) = (3cos(-165°),3sin(-165°)) = (-3cos15°,-3sin15°)
So you have the right idea, but you need to translate between the two orientations.
All angles are measured CCW from +x-axis.
X = 3*Cos195+4*Cos282 = -2.07 N.
Y = 3*sin195+4*sin282 = -4.69 N.
Fr = sqrt((-2.07)^2 + (-4.69)^2 = 5.13 N. = Resultant force.
X = 3*Cos195+4*Cos282 = -2.07 N.
Y = 3*sin195+4*sin282 = -4.69 N.
Fr = sqrt((-2.07)^2 + (-4.69)^2 = 5.13 N. = Resultant force.
2 answers