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

Confusion regarding resolution of forces

  • #1

Homework Statement



A force which is at an angle θ to the horizontal (not being 90°) has two components Fx = FCosθ and Fy = Fsinθ along the x and y direction. However, when working out problems regarding objects on an incline, the horizontal (parallel to the slope) component of Normal force is taken as mgSinθ and the vertical (perpendicular to the slope) component is taken as mgCosθ. Can somebody explain ? I know this sounds silly, but I am confused.

Homework Equations





The Attempt at a Solution



Is it something to do with the CCW convention?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,427
1,824
Try to draw your problem. In the first case, you have a force that makes an angle with the horizontal direction.

In the second case, you have the normal force, that is perpendicular to the slope. The slope makes an angle theta with the horizontal. What angle does the normal force enclose with the horizontal?

ehild
 
  • #3
The slope makes an angle θ to the horizontal and the perpendicular to the slope makes an angle 180 - (90 + θ). So if the inclination of the slope is 30°, then the perpendicular makes an angle 60° to the horizontal surface. So whatever angle it makes with the horizontal, the angle made by the perpendicular with the horizontal is 90-θ. Therefore the perpendicular component is mgSinθ. Is my reasoning correct?
 
  • #4
ehild
Homework Helper
15,427
1,824
Well, the normal force is normal to the slope, so its perpendicular component is itself, and the parallel component is zero. Do you meant gravity, mg? It is vertical and has components parallel and perpendicular with respect to the slope. See attachment: what are the x and y components of the vertical force F?


ehild
 

Attachments

Last edited:
  • #5
The force F is resolved into x and y components (I did mean gravity). But I still don't get it. I know that sinθ = cos(θ-90). But θ is the angle of the slope and how does this angle come into the picture when resolving for F? I have looked around but I am not able to find an answer. Probably because this is so basic.
 
  • #6
ehild
Homework Helper
15,427
1,824
Yes, it is so basic. The blue angle is theta. What is the green one in the yellow triangle?

ehild
 

Attachments

  • #7
90 - θ ? I don't think that this angle can be equal to θ.
 
  • #8
ehild
Homework Helper
15,427
1,824
90 - θ ? I don't think that this angle can be equal to θ.
Yes, it is 90 - θ ! So what are the components (parallel with the slope and perpendicular to the slope) of the vertical force mg?


ehild
 
  • #9
Then the perpendicular component should be mg cosθ and the parallel component is mg sinθ? I get it now. Thanks for your patience ehild.
 
  • #10
ehild
Homework Helper
15,427
1,824
You mean that component of the vertical force, which is perpendicular to the slope. Is it right? Look at the yellow triangle. The vertical component is the side opposite to the green angle. So it is Fsin(green angle) but the green angle is 90-θ. sin(90-θ)=cos(θ).

ehild
 

Related Threads on Confusion regarding resolution of forces

  • Last Post
2
Replies
42
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
5
Views
490
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
3
Views
761
Top