1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confusion regarding resolution of forces

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



    3. The attempt at a solution

    Is it something to do with the CCW convention?
     
  2. jcsd
  3. Oct 25, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  4. Oct 25, 2011 #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?
     
  5. Oct 25, 2011 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     

    Attached Files:

    Last edited: Oct 25, 2011
  6. Oct 25, 2011 #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.
     
  7. Oct 26, 2011 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, it is so basic. The blue angle is theta. What is the green one in the yellow triangle?

    ehild
     

    Attached Files:

  8. Oct 26, 2011 #7
    90 - θ ? I don't think that this angle can be equal to θ.
     
  9. Oct 26, 2011 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  10. Oct 26, 2011 #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.
     
  11. Oct 26, 2011 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Confusion regarding resolution of forces
  1. Resolution of Forces (Replies: 9)

  2. Resolution Of Forces (Replies: 1)

  3. Force Resolution (Replies: 2)

Loading...