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undertow2005
Jun7-04, 03:11 PM
heres the problem

represent on a diagram the following to vectors:1) 3m, 45 degrees north of east; and 2) 6m, 60 deg west of north. find the resultant of sum of these two vectorsin terms of magnitude and angle with respect to the east.


nevermind the first part of the Q, i need to find the second part


my incorrect solution:

for vector one:
Cx=3m*cos45=2.121
Cy=3*sin45=2.121

Vector 1=2.121i + 2.121j

Vector two:
Cx=6*cos60=3
Cy=6*sin60=5.196

vector 2= 3i +5.196

i added them and got this 5.121i +7.317j

someone check!

My Final only two hours away!

BobG
Jun7-04, 03:35 PM
You've got the mechanics of it down, but you're not paying enough attention to your principle directions. These vectors aren't referenced to the same principle direction.

The first is referenced to East - your components were correct.

The second is referenced to North - you need to add 90 degrees to the angle to convert it to an East reference (you also made a mistake by using positive 60, which is 60 degrees East of North, but that's irrelevant since you started from the wrong reference).

The 'i' of your second vector is the cosine of 150 degrees - the 'j' the sine of 150 degrees. (the cosine of 150 is negative, meaning your resultant will be a positive plus a negative).

undertow2005
Jun7-04, 03:46 PM
You've got the mechanics of it down, but you're not paying enough attention to your principle directions. These vectors aren't referenced to the same principle direction.

The first is referenced to East - your components were correct.

The second is referenced to North - you need to add 90 degrees to the angle to convert it to an East reference (you also made a mistake by using positive 60, which is 60 degrees East of North, but that's irrelevant since you started from the wrong reference).

The 'i' of your second vector is the cosine of 150 degrees - the 'j' the sine of 150 degrees. (the cosine of 150 is negative, meaning your resultant will be a positive plus a negative).

vector 2

6*cos(-150=-5.196
6*sin(-150=-3.00

correct?

BobG
Jun7-04, 05:09 PM
Almost. Sine of 150 is positive one-half (+0.5).

Sine is positive from 0 to 180

Cosine is positive from 270 to 90

If you plot this on a graph, your cosine is your x and your sine is your y, which makes it pretty easy to see which quadrants are positive or negative.