# Thread: Need to find G on a ramp

1. ## Need to find G on a ramp

I have a block of ice sliding down a FRICTIONLESS ramp with an accelleration of 5.3 m/s.s.. If the ramp makes an angle of 32.7 degrees with the horizontal , find g, the acceleration due to gravity.

This is using the application of vectors, the only way that I can solve this.
But I am not sure. what to do, help get me started please.

2. Draw a figure. Always when you have a problem in classical kinematics and kinetics, draw a figure. There's a reason every single textbook on classical mechanics include a section about free body diagrams.

In this case, you have an inclining plane and a block on it. There are two forces involved - identify which they are and add them into the figure in the appropriate place. Gravity acts at the center and so on.
Also, don't forget to specify a set of coordinates. The convention for this type of problem is to have the positive x-direction in the direction of motion along the incline of the plane, and the positive y-direction then becomes in the direction of the normal force from the plane on the block.
After that it's application of Newton's second, F = ma all the way.

3. How do I do that calculation if all I have is a degree and an Acceleration of 5.3 m/s.s.
That is what I am not sure about??

How do I do that calculation if all I have is a degree and an Acceleration of 5.3 m/s.s.
That is what I am not sure about??
what is Newton's 2nd law of motion and what does it mean?

5. well it's F=ma. Sorry still lost on this one..
Do I take 5.3 and / by 9.8 to find the velocity?
Which is -5.4 m/s.

6. if you look at the force diagram, the normal force , $\displaystyle N$ , is equal and opposite to the weight component perpendicular to the incline, $\displaystyle mg\cos{\theta}$ ... in essence, those two forces cancel.

... which leaves the parallel component of weight, $\displaystyle mg\sin{\theta}$ , alone as the net force.

2nd law ...

$\displaystyle F_{net} = ma$

$\displaystyle mg \sin{\theta} = ma$

$\displaystyle g\sin{\theta} = a$

... the standard formula for the acceleration of an object down a frictionless incline with elevation angle $\displaystyle \theta$.

7. So is the answer 9.81 g??
5.3/sin32.7 ?

8. The answer is 9.81 m/s^2, not 9.81 g, lol.

9. Well I have to find g. So why would it be 9.81m/s.s.?? if it's the weight?
Or do I have it confused somehow?

On never mind, re read the question, and its g, the acc due to gravity.
GOT IT, thanks.