• Support PF! Buy your school textbooks, materials and every day products Here!

Vector magnitude and direction

  • Thread starter magnifik
  • Start date
  • #1
360
0

Homework Statement


Find the magnitude of the resultant force and the angle it makes with the positive x-axis

draw one vector from origin (0,0) in quadrant one that is 45 degrees (this is 20 lbs)

draw another vector from the origin (0,0) in quadrant four that is 30 degrees (this is 16 lbs)


The Attempt at a Solution


here's what i did initially
20cos(45)i + 20sin(45)j = 10*sqrt(2)i + 10*sqrt(2)j
16cos(-30)i + 16sin(-30)j = 8*sqrt(3)i - 8j
F = (10*sqrt(2)i + 10*sqrt(2)j) + (8*sqrt(3)i - 8j)
|F| = sqrt((10*sqrt(2)i + 10*sqrt(2)j)^2) + (8*sqrt(3)i - 8j)^2)) = 28.66 N
[tex]\vartheta[/tex] = (10*sqrt(2) - 8)/(10*sqrt(2) + 8*sqrt(3)) = 12.37 deg

but if you use cosine rule or sine rule, you get different answers...
The angle is 45* + 30* = 75*

Now we use the Cosine rule to find the magnitude of the resultant vector. The resultant vector is 22.1 lbs.

Use the Sin rule and the angle is 90.7* (basically 90*)

which one is right?
 

Answers and Replies

  • #2
9
0
The angle of the resultant vector should be between the angles of your initial vectors. Your answer of 12.37 degrees seems reasonable...however, 75 degrees and 90.7 degrees do not.
 

Related Threads on Vector magnitude and direction

Replies
3
Views
1K
Replies
0
Views
908
Replies
5
Views
4K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
6K
  • Last Post
Replies
3
Views
1K
Top