Free-Body Diagram

  • Thread starter amcavoy
  • Start date
  • #1
amcavoy
665
0
I was trying to figure out how to get the outward normal vector to the surface of a ramp inclined θ degrees from the horizontal. Say that a block of mass "m" is on the surface and the surface is frictionless. When I draw the free-body diagram, I come up with a downward force of <0,-mg>. To calculate the force in the direction of the incline, I first want to find the normal vector to add to <0,-mg>. Is this the correct way to do this? It looks like |n|=|g|cosθ, but I cannot find the coordinates of this vector. Any ideas?

Edit: Working it out, I came up with the following for the coordinates of the outward normal vector:

x = |g|cosθsinθ
y = |g|cos2θ

Are these correct?
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
45,465
1,956
You got it--except for a missing mass (you left out the "m" in your equations). (Note that it's often more useful to use coordinates parallel and perpendicular to the incline surface, rather than vertical and horizontal components.)
 
Last edited:
  • #3
amcavoy
665
0
Oh yeah, the m! Anyways, thanks for the response :smile:.
 

Suggested for: Free-Body Diagram

Replies
4
Views
398
Replies
3
Views
938
Replies
6
Views
379
Replies
9
Views
667
Replies
13
Views
2K
Replies
14
Views
1K
Replies
1
Views
695
  • Last Post
Replies
15
Views
439
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
7
Views
895
Top