# Vector Problem

• September 3rd 2009, 01:44 PM
ur5pointos2slo
Vector Problem
Could someone please verify if this problem is done correctly please. If not where did I go wrong.

The problem:

Use the component method to find the resultant(magnitude and direction) of the forces listed below.

A 200 N at 20 degrees N of East
B 250 N at 180 deg.
C 500 N at -150 deg. S of East

My attempt at the solution:

Ax = 500 cos (-150) = - 433
Ay = 500 sin (-150) = -250
Bx = -250
By = 0
Cx = 200 cos 20 = 187.9
Cy = 200 sin 20 = 68.4

R = (-433-250+187.9)I + (-250 + 0 + 68.4)j
= (-495.1, -181.6)
R = sqrt(-495.1^2 + -181.6^2)
= 527

Tan Theta = 318.4 / 370.9
Theta = 20.1 deg.
• September 3rd 2009, 01:49 PM
Matt Westwood
Apart from getting A and C the wrong way round ... are you sure that 180 degrees is due west and not due south?
• September 3rd 2009, 01:51 PM
ur5pointos2slo
Quote:

Originally Posted by Matt Westwood
Apart from getting A and C the wrong way round ... are you sure that 180 degrees is due west and not due south?

Yes it is supposed to be due west.
• September 3rd 2009, 01:53 PM
Matt Westwood
Sorry, it's just an incredibly unnecessarily badly worded problem.
• September 3rd 2009, 02:00 PM
ur5pointos2slo
Quote:

Originally Posted by Matt Westwood
Sorry, it's just an incredibly unnecessarily badly worded problem.

Maybe I should draw a diagram. I think that will help.
• September 3rd 2009, 02:04 PM
ur5pointos2slo
• September 6th 2009, 09:41 AM
ur5pointos2slo
Noone can tell me if this problem looks correct? Using the diagram given use the component method to determine the resultant(magnitude and direction). Forget how it was worded before, now all you have to do is look at the diagram.
• September 6th 2009, 10:34 AM
skeeter
$R_x = 200\cos(20) + 250\cos(180) + 500\cos(-150)$

$R_y = 200\sin(20) + 250\sin(180) + 500\sin(-150)
$

$|R| = \sqrt{R_x^2 + R_y^2}$

since the resultant components end up in quad III ...

$\theta = \arctan\left(\frac{R_y}{R_x}\right) + 180$

where $\theta$ = angle measured from due east (the + x-axis)
• September 6th 2009, 10:56 AM
ur5pointos2slo
Quote:

Originally Posted by skeeter
$R_x = 200\cos(20) + 250\cos(180) + 500\cos(-150)$

$R_y = 200\sin(20) + 250\sin(180) + 500\sin(-150)
$

$|R| = \sqrt{R_x^2 + R_y^2}$

since the resultant components end up in quad III ...

$\theta = \arctan\left(\frac{R_y}{R_x}\right) + 180$

where $\theta$ = angle measured from due east (the + x-axis)

Thank you very much!