Magnitude and direction angle

  • Thread starter freespirit
  • Start date
  • #1
My problem is to determine the magnitude and direction of the resultant force FR=F1+F2 and it's direction, measured counterclockwise from the positive x direction.

f1=250 lb @ 60 degrees from x
f2= 375 lb @ -45 degrees from x

Ok I got the magnitude by doing this:
(360-2(255))/2=-75 degrees

fr=sqroot of (250^2+375^2-2(250)(375)cos(75)
fr=393.188~ 393

then I got the angle by this:
375/sin x = 393.188/sin 75
x=67.1088
how do i get the resultant angle, what do I need to add to the 67 degrees?
 

Answers and Replies

  • #2
nautica
breakdown x and y components of each by the following

X= r cos(angle)
Y= r sin(angle)

Add the two x and y compents to get the x and y of the resultant. Use the pathagorean theroum to get the resultant magnitude and use tan^(-1) (y/x) to get resultant directions.

nautica
 
  • #3
650
1
Preview Rough fig in attachment so

[tex] tan{\alpha} = \frac{ysin{\theta}}{x+ycos{\theta}}[/tex]

[tex] tan{\alpha} = \frac{ysin{\theta}}{x+ycos{\theta}}[/tex]
 

Attachments

  • untitled.jpg
    untitled.jpg
    9.9 KB · Views: 573
Last edited:
  • #4
Thank You

Thank you both for your help.
 

Related Threads on Magnitude and direction angle

Replies
2
Views
3K
Replies
1
Views
2K
Replies
12
Views
27K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
4
Views
4K
Top