Lets say there is an object on a ramp of 45 degrees. There is no friction between the object and the ramp. Will the object have the same acceleration as an object in free falling?(g)

 

 

 

 

 

 

 

 

Well, you can draw out a force diagram to see this one. In free-fall, there's no force opposing the fall. When an object slides down the ramp, is there a force pushing back?

Well there is, but the no friction part makes me confused.

Well there is, but the no friction part makes me confused.

 

 

If you drop the object it falls down.

 

If you put it on a frictionless ramp, it follows the ramp.

 

Acceleration is a vector -- there is a direction associated with it.

 

Note also that the object on the ramp travels a greater distance than the one that is simply dropped, before reaching the ground. Now ask yourself what conservation of energy tells you about the final speed. Would you expect the magnitude of the acceleration to be the same in both cases ?

I still can't think of a way to calculate that. Maybe use Pythagorean theorem .. but how exactly idk.

 

Maybe since the acceleration i'm interested in is the hypotenuse of the triangle...

 

[math]x = \sqrt{( sin(45 deg) * g ) ^2 + (cos(45 deg) * 0) ^2 }[/math]

 

[math]x = \sqrt{\frac{1}{2} * g^2 + 0 }[/math]

 

[math]x = \frac{1}{\sqrt{2}} * g[/math]

 

 

 

 

 

 

 

 

Am i close?

Edited October 31, 201114 yr by deemeetar

Hypotenuse of which triangle?

 

You can break down the gravitational force into components: a component perpendicular to the surface of the ramp, and one parallel to it. It's the parallel one that will cause the object to slide down the ramp. Will that accelerate it the same as if there were no ramp?

Well my imaginary triangle was from the perpendicular(horizontal) and the parallel force(vertical). I guess since the move has different direction that the acceleration will be different but how much different. I know the perpendicular force is 0, the parallel one is G = mg. I just can't make a connection to the one under angle of 45 deg. Please tell me how to calculate that? What should i use? Am i close with calculating it like i was trying or i am far from it?

There are components parallel and perpendicular to the surface of the ramp. You want to split the gravitational acceleration into those components. Consider this diagram, if you ignore F_a and F_f (since there's no friction):

 

 

Clearly the component of gravity pushing the block straight into the ramp won't accelerate it, because the ramp pushes back. The component pushing down the ramp, however, will accelerate it. You can use some trig, knowing [math]\Theta[/math] and F_g, to find that component.

Oh thank you. I've imagined it a lot more different.

 

So here's my solution:

 

[math]sin(45deg) = \frac{Fg2}{Fg}[/math]

 

[math]Fg2= \frac{\sqrt{2} * Fg}{2}[/math]

 

 

[math]m * a = \frac{\sqrt{2} * m * g}{2} [/math]

 

[math]a = \frac{\sqrt{2} * g}{2} [/math]

Yup. And is that acceleration greater than or less than g?

Less than g