What is the Acceleration of Gravity on an Inclined Plane?

AI Thread Summary
The discussion focuses on deriving the gravitational acceleration on an inclined plane, expressed as g = g₀sin(θ)êₓ - g₀cos(θ)êᵧ, where g₀ equals 9.81 m/s². Participants explore the trigonometric relationships involved, noting that sin(θ) corresponds to the component of gravity acting along the incline, while cos(θ) relates to the perpendicular component. Confusion arises regarding the signs of these components, with clarification provided on the positive x-direction and negative y-direction. Visual aids and diagrams are suggested to better understand the vector components and their relationships. The conversation emphasizes the importance of correctly applying trigonometric principles to resolve the components of gravitational force on an incline.
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Homework Statement



[PLAIN]http://img192.imageshack.us/img192/8160/3233q.jpg

Need to show the gravitational acceleration:

g = g_{0}sin(\theta)\hat{x}-g_{0}cos(\theta)\hat{y}

where g_{0}=9.81ms^{-2}.

(note that the g, x and y, should all have underbar notations).

Homework Equations



[included within problem statement and solution attempt]

The Attempt at a Solution



I know this is simple and easy but I seem to have just forgotten how to do trigonometry! :blushing:
 
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What is the problem? As far as I can see, all you would have to do is plug in the angles, if there is data on it, and to find the magnitude, you would use the pythagorean theorem.
 
IBY said:
What is the problem?

The diagram is all the information that I am given, and I need to show g.

IBY said:
As far as I can see, all you would have to do is plug in the angles, if there is data on it, and to find the magnitude, you would use the pythagorean theorem.

Yes, but I can't figure it out [mindblock] :redface:.
 
[PLAIN]http://img249.imageshack.us/img249/8338/3233q2.jpg

OK so perhaps the simple question should now be.. why is it just the sin component minus the cos component?

.. sorry for sounding stupid about this ..
 
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Sin is opp/hyp, and in this case, opposite is the axis in which the box slides. Hypotenuse is g. If you draw a line between the arrows of g*sin(theta) and g, it forms a right triangle. Theta, in that triangle, is in the bottom left corner because the upper left forms 90-theta, and the right corver is 90. Use the same kind of reasoning with g*cos(theta).

The reason it is minus, by the way, is easy. Which way is the +x direction, and which way is the +y direction?
 
IBY said:
Sin is opp/hyp, and in this case, opposite is the axis in which the box slides. Hypotenuse is g. If you draw a line between the arrows of g*sin(theta) and g, it forms a right triangle. Theta, in that triangle, is in the bottom left corner because the upper left forms 90-theta, and the right corver is 90. Use the same kind of reasoning with g*cos(theta).

OK so have I understood that correctly? (see diagram and calcs)

[PLAIN]http://img35.imageshack.us/img35/9889/323qf.jpg

This means:

sin{\theta} = \frac{gsin{\theta}\hat{x}}{g}

sin{\theta} = \frac{-gcos{\theta}\hat{y}}{g}

.. but then don't see where I go from here (if that's even correct).

IBY said:
The reason it is minus, by the way, is easy. Which way is the +x direction, and which way is the +y direction?

yes since taking negative y component and positive x component.
 
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No, on the y component, theta is on the right up, not right down. You got the 90 degrees right. Think about it, the shape of the area of vector x and y is a rectangle. All angles of a rectangle is 90 degrees, which means that if you split the rectangle with a diagonal, you see that the two angers at one corner of the rectangle are 90 degrees. Which means that the other angle is 90-theta. I think it would be helpful if you stuck the two drawings you made above together where g meets for visualization. You will see that it makes no sense for there to be two thetas at the bottom corner.
 
[PLAIN]http://img827.imageshack.us/img827/6956/2324q.jpg
 
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