Body's Acceleration: Calculating Acceleration on Fixed Platform

[willydavidjr]

I have a question in my mind. If the body is supposed to slide on with mass m on a fixed triangular platform with angle \theta. Ignoring friction and with g as gravitational acceleration, what will be the acceleration of the body if it slide on the platform?

 

[Doc Al]

Consider the forces acting on the mass parallel to the surface (since that's the only direction the mass can move).

 

[willydavidjr]

In short, the acceleration of the body is the same as the gravitational pull of the earth(9.8 m/s^2)disregarding the friction on the platform?How about the angle?Do I need to disregard it?

 

[Hootenanny]

Do you mean there is a body on an incline? As in this image;

http://upload.wikimedia.org/wikipedia/en/e/e7/Free_body.gif

~H

 

[Doc Al]

In short, the acceleration of the body is the same as the gravitational pull of the earth(9.8 m/s^2)disregarding the friction on the platform?

9.8 m/s^2 is the acceleration of a freely falling body (neglecting air resistance)--it is not the acceleration of an object sliding down an incline.

How about the angle?Do I need to disregard it?

Most certainly not; the angle is key. As I had asked before, identify the forces acting parallel to the surface. (Hoot gave you a good diagram to refer to.) Then you can apply Newton's 2nd law to find the acceleration.

 

[willydavidjr]

Meaning the acceleration of the body is -mgsin\theta by disregarding the friction of the platform. My answer would be negative because it is going downward?

 

[Hootenanny]

Meaning the acceleration of the body is -mgsin\theta

That would be the force acting down the slope. Simply apply Newton's law, F=ma to find the acceleration.

~H

edit: sorry for jumping in on you Doc, I didn't realize you were still online.

 

[Doc Al]

Meaning the acceleration of the body is -mgsin\theta by disregarding the friction of the platform.

That's the force, not the acceleration. (What is the acceleration?)

My answer would be negative because it is going downward?

The acceleration is down the incline. Depending upon the sign convention used, that could be negative. Perhaps you only need the magnitude of the acceleration.

 

[willydavidjr]

So the acceleration of the body from Newton's law, all forces from x-axis would be -mgsin\theta=ma_x if we disregard friction. So the magnitude of the acceleration is a=gsin\theta. Am i correct now?

 

[Doc Al]

Now you've got it.

(No problem, Hoot. The more, the merrier! )

 

[willydavidjr]

This question would be interesting. What if the platform moves with an acceleration of a on the horizontal plane, the body stands still on the slope of the moving stand. What is the acceleration a of the stand?

 

[Hootenanny]

This question would be interesting. What if the platform moves with an acceleration of a on the horizontal plane, the body stands still on the slope of the moving stand. What is the acceleration a of the stand?

Can you figure it out yourself? What would be the forces involed? Direction is important for this question.

~H

 

[Doc Al]

You'd attack this problem by identifying all the forces on the body, then applying Newton's 2nd law. In this case you'd be solving for the conditions which make the vertical acceleration zero. (Write separate equations for the horizontal and vertical components.)

 

[willydavidjr]

I am sorry because I still catching up on these forces acting upon the body.

 

[willydavidjr]

I want to quote the image. Should it be -mgcos\theta?