Accelertion Due to gravity on a ramp equation

  • Thread starter Thread starter krsbuilt
  • Start date Start date
  • Tags Tags
    Gravity Ramp
Click For Summary

Homework Help Overview

The discussion revolves around the acceleration due to gravity on an inclined ramp, focusing on the relationship between the angle of the ramp and the resulting acceleration in the direction of the ramp. Participants are analyzing various equations and concepts related to gravitational force and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive an equation relating the angle of the ramp to acceleration, while others question the validity of this approach and suggest using trigonometric relationships instead. Some participants discuss breaking down forces into components and the implications of mass on acceleration.

Discussion Status

Participants are actively engaging with the problem, with some offering alternative equations and insights into the physics involved. There is a recognition of the need to reconsider the original equation and the effects of different objects on acceleration.

Contextual Notes

There are mentions of potential inaccuracies in the original equation due to assumptions made about the relationship between angle and acceleration. The discussion also touches on the impact of using different objects, such as a ball versus a cart, in experiments.

krsbuilt
Messages
15
Reaction score
0
I'm new to this forum so please don't mind my being a newbie. I found this equation at school, .5((90-theta)/90)*G=Ag where theta is the angle on the base (not the right angle)


|\
| \ <ramp
|_ \<Theta
^
right angle

and G is the acceleration due and Ag is acceleration in the direction of the ramp. i found this bu knowing if it's a 90 (horizontal/flat) degree angle then there is no acceleration and if it's 0 (vertical) degrees then there's full acceleration. so every degree in between 0 and 90 it adds (1/2)1/90*G to the acceleration to the acceleration in the direction of the ramp.
 
Physics news on Phys.org
Your equation is incorrect.

While it may seem tempting at times to contrive some relationship between two data points an extrapolate a linear relationship (what you've done), this type of analysis is often incorrect (and difficult to prove).

The correct equation results from the fact that the gravitational force parallel to the incline is given by:
F_{//}=mgsin(\theta)

So acceleration is given by:
a_{//}=gsin(\theta)
 
you have to break it up into two components. Thats the idea and that's what makes it a triangle and for triangles we have trigonometry. So how much "down" are you going if you are going that much "side"? well Fg=mg but that's only if you drop something directly down to the ground so if its on an angle and its the angle of the slope to the gound then you need to do (mg)sin(theta). So this gives you the force, now you have to find the acceleration and f=ma so f/m=a so just divide your previous answer by m and you are left with a. Hey what do you know? Newton's observations were correct, mass is canceled and irrelevant because the object accelerates at a certain rate despite it's mass. woohoo! i took a few extra steps to show you how the idea works :P. Oh by the way, a sin function is nonlinear therefore your linear equation i.e. (1/2)1/90*G simply can't be right. i mean i could be missing something but the only time lines and curvy lines are related and interchangeable is in calculus and this is high school physics so unless your prof or someone used calculus to get this relationship i just don't think it makes sense. The rate of change changes depending on theta so you can't treat it as if it doesn't and expect to be right.
 
i decided to change the equation to theta/90*g=Ag... My physics teacher told me that it was sheer luck that my data came close to the same. i was using a ball down a ramp and apparently the ball rolling can cause changes in the accelaration. so I'm going to change the ball to a cart and see what went wrong so i can figure out that the equation is actually G*sin(theta)=Ag.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 56 ·
2
Replies
56
Views
5K
Replies
17
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
4K