Finding Weight on Inclined Plane

[tod88]

EDIT: I think I just figured it out...could you check out the picture I added and see if my reasoning is right?

Thanks!

http://img141.imageshack.us/img141/618/20080716103213jm3.png

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I'm working on free-body diagrams and I'm having trouble finding the components of the weight of an object on an inclined plane. I know that if you take the +x axis to be the plane and the +y axis to be the direction of the normal force that the weight components are:

W parallel = mgsin(theta)
W perpendicular = mgcos(theta)

where theta is the angle of inclination

My problem is that I have trouble "seeing" geometrically how to actually FIND the components using trig functions. Is the weight -always- going to be determined by these two equations if I use the coordinate system that I mentioned?

Sorry if this is too basic a question! I just wanted to see the "why" behind these equations.

 

[PhanthomJay]

Your angles are shown incorrectly. What you have shown as (90- A) is actually A, and vice versa. Carefully do the geometry first, then the trig. You will find that the weight component parallel to the incline is mgsinA, and the weight component perpendicular to the incline is mg cosA.

 

[alphysicist]

Hi tod88,

EDIT: I think I just figured it out...could you check out the picture I added and see if my reasoning is right?

Thanks!

http://img141.imageshack.us/img141/618/20080716103213jm3.png

------------
I'm working on free-body diagrams and I'm having trouble finding the components of the weight of an object on an inclined plane. I know that if you take the +x axis to be the plane and the +y axis to be the direction of the normal force that the weight components are:

W parallel = mgsin(theta)
W perpendicular = mgcos(theta)

where theta is the angle of inclination

My problem is that I have trouble "seeing" geometrically how to actually FIND the components using trig functions. Is the weight -always- going to be determined by these two equations if I use the coordinate system that I mentioned?

Sorry if this is too basic a question! I just wanted to see the "why" behind these equations.

I notice you have one set of equations in the diagram, and a different set in the post itself. (In your picture, you say the cosine function is associated with the parallel component, and in the post itself you say the cosine function goes with the perpendicular component.)

When the box is just on level, flat ground, that corresponds to your angle of inclination A (or \theta) being zero degrees. In that situation, is the perpendicular or parallel component of the weight equal to zero? (parallel or perpendicular to the ground) Which set (in the picture or in the post) of equations agrees with that?

 

[tod88]

Ah...that makes sense. So whenever there's no incline, all the weight is perpendicular (mg)(cos(0)) = mg and whenever the incline is 90˚, all the weight is parallel (mg)(sin(90˚) = mg and the object is in free-fall.

And thanks for the comment on my diagram too...I see where I messed up the angles. I think I understand now.