Is the acceleration constant?

1. Dec 3, 2008

wallpaper

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

Is the acceleration constant? Why?

2. Dec 3, 2008

Mentallic

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.

3. Dec 3, 2008

wallpaper

Ooh thanks. I have another question though, how does the degree of the tilt affect the acceleration?

4. Dec 3, 2008

Troels

Like it was said:

5. Dec 3, 2008

Mentallic

I'll be a little more clear here

What you need to do is draw a diagram of the situation, labelling the angle $$\theta$$ 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:

$$sin\theta=\frac{opp}{hyp}$$

$$sin\theta=\frac{g}{a_f}$$
where:
af being the accleration of the cart
g=acceleration due to gravity

Therefore, $$a_f=\frac{g}{sin\theta}$$

6. Dec 3, 2008

nasu

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: Dec 3, 2008
7. Dec 3, 2008

Mentallic

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?

8. Dec 3, 2008

Staff: Mentor

What you are trying to find is the component of g parallel to the incline, so g should be the hypotenuse of your triangle.