Is Acceleration Constant on an Inclined Plane?

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Discussion Overview

The discussion revolves around the question of whether acceleration is constant for a cart rolling down an inclined plane. Participants explore the implications of incline angle and the effects of gravity on acceleration, considering both ideal conditions and potential complications.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that acceleration is constant if the incline angle and gravitational force are constant, assuming no resistances.
  • Others inquire about how the angle of the incline affects the acceleration, suggesting that the acceleration vector is related to the sine of the angle.
  • One participant proposes a method to visualize the problem using a diagram to illustrate the relationship between the angle, gravity, and acceleration.
  • There is a discussion about the correct mathematical representation of the acceleration, with some confusion regarding the relationship between the sine of the angle and the acceleration of the cart.
  • Concerns are raised about the implications of using incorrect formulas, such as suggesting that using g/sin(theta) would lead to infinite acceleration on a horizontal plane.

Areas of Agreement / Disagreement

Participants express differing views on the mathematical relationships involved and the implications of the incline angle on acceleration. There is no consensus on the correct interpretation of the trigonometric relationships or the resulting acceleration values.

Contextual Notes

Some participants note confusion regarding the application of trigonometric principles to the problem, indicating potential misunderstandings in the mathematical reasoning. The discussion does not resolve these confusions.

wallpaper
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Consider the system of a cart rolling down an inclined plane.

Is the acceleration constant? Why?
 
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Yes the acceleration is constant assuming no resistances and the incline is at a constant angle.

The acceleration vector of the cart is equal to the sine of the angle to the horizontal made by the incline. Since the angle and gravity are constant, the acceleration of the cart is also constant.
My reasoning as to why may only be intuitive though.
 
Ooh thanks. I have another question though, how does the degree of the tilt affect the acceleration?
 
wallpaper said:
Ooh thanks. I have another question though, how does the degree of the tilt affect the acceleration?

Like it was said:

Mentallic said:
The acceleration vector of the cart is equal to the sine of the angle to the horizontal made by the incline.
 
wallpaper said:
Ooh thanks. I have another question though, how does the degree of the tilt affect the acceleration?

I'll be a little more clear here :smile:

What you need to do is draw a diagram of the situation, labelling the angle [tex]\theta[/tex] and the gravity acceleration vector (pointed directly towards the floor with a value of 9.8ms-2 if this scenario is on Earth's surface).
Now, what you need to find is the acceleration vector of the cart. This is simply the hypotenuse of your right-angled triangle that you were able to illustrate with the vectors. Since you have the angle, the opposite side and the hypotenuse; the relationship is:

[tex]sin\theta=\frac{opp}{hyp}[/tex]

[tex]sin\theta=\frac{g}{a_f}[/tex]
where:
af being the acceleration of the cart
g=acceleration due to gravity

Therefore, [tex]a_f=\frac{g}{sin\theta}[/tex]
 
The acceleration is

g*sin(theta)

where theta is the angle of the incline.
This is assuming no friction, of course.
If it were g/sin(theta) you'll get infinite acceleration on a horizontal plane...
 
Last edited:
Yes your logical thinking sounds much more correct. However, this simple trigonometry is confusing me because I can't get that result. Can anyone spot my mistake?
 
Mentallic said:
Now, what you need to find is the acceleration vector of the cart. This is simply the hypotenuse of your right-angled triangle that you were able to illustrate with the vectors.
What you are trying to find is the component of g parallel to the incline, so g should be the hypotenuse of your triangle.
 

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