Homework Help: Box and Ramp problem

1. Aug 29, 2004

mrjeffy321

Lets say you have a box at the bottom of a frictionless ramp and you give it a velocity (X) of lets say 10 m/s, so that when it gets on the ramp, the components velocities would be:
Vx = 10 * cos (angle)
Vy = 10 * sin (angle)

then once it gets ontp the ramp, the gravitational acceleration changes from
being -9.81 m/s^2 down (Y) to a combination of X and Y acceleration.

I am trying to find the X and Y acceleration, I have been looking at the diagram of the ramp and box and have drawn my force lines to indicate which direction everthing has a force exserted on it, but when I goto calculate the acceleration I get very confused.

So lets just say in this example the ramp is at an angle of 50 degrees, the box has in initial X velocity (before the ramp) of 10 m/s and we will esimate gravitational acceleration at -10 m/s^2.
so we now have the box on the ramp with:
Vx = 10 * cos (60) = 5 m/s
Vy = 10 * sin (60) = 8.7 m/s

acceleration X = ? -10 * sin (60) = -8.7 m/s^2 ?
acceleration Y = ? -10 * cos (60) = -5 m/s^2 ?

is that right? or is it:

acceleration Y = ? -10 * sin (60) = -8.7 m/s^2 ?
acceleration X = ? -10 * cos (60) = -5 m/s^2 ?

2. Aug 29, 2004

Gmaximus

Define your coordinate system as X is parrallel to the ramp and Y is parrallel to the normal force.

Then: F=ma and F due to grav is mg.
I think you might be confused because your box is massless. You have to sum your forces, dont try adding accelerations.

3. Aug 29, 2004

mrjeffy321

OK, I get that now,

so to find the acceleration in a particular direction use the opposite (sin/cos) finction that you used to find its velocity in that difrection.
so it would be:
acceleration X = ? -10 * sin (60) = -8.7 m/s^2 ?
acceleration Y = ? -10 * cos (60) = -5 m/s^2 ?

this is what I was afraid of, that just caused more problem than it solved, because in the long run of this problem, I am trying to find the angle of the ramp, and the distance traveled in each direction (X and Y).

4. Aug 29, 2004

Gmaximus

I don't think this can be done without a mass of the box.

Do you have the actual text of the problem?

5. Aug 29, 2004

mrjeffy321

No, the mass of the box is not needed in the problem, im pretty sure of that (it isnt give anyway). I do have the actual text of the problem, an I can give it to you if you want, but I was trying to maintain a certain degree of integrity with this by doing it myself.

I will tell you what it wants:
basicaly, the problem gives you a box at a ramp with an initial horizontal velocity (which you then convert to its componet velocity), and it gives you a distance traveled (X). it then asks you to find the angle of the ramp and the distance traveled (Y) and the time it takes.

I assume that I can use the formula: Vf = Vi + a * t, to find the time, and thus the distance traved along with this formula, d = (Vi * t) + (1/2*a*t^2).
but then I ran accross this other way of finding the angle:
angle = tan^-1 (Force applied / Force gravity)
and you can get the force applied by using the formula:
Vf^2 = Vi^2 + 2 * a * d

6. Aug 29, 2004

Gmaximus

But the magnitude of the acceleration due to gravity is directly related to the mass...

Well then, im confused. Oh well!

7. Aug 29, 2004

mrjeffy321

o great, now I have spreak my confusion to the only other person who was helping me.

the magnitude of acceleration is related to the angle of the slope. the magnitude of the force is realated to the mass. but in this problem, I dont care about force.

Here is what I got for an answer, but I am almost certain that it is wrong.
Dx = 5.1 m
Dy = 8.6 m
angle = 59 degrees
time = 1 second up, 1 second down

I say that it is almost certainly wrong because I got that back when I still wasnt sure how to correctly calculate the component accelerations, and I thing I did it wrong.

8. Aug 30, 2004

HallsofIvy

Yes, you are very confused. The acceleration due to gravity on an object is independent of the object's mass!

(The force is proportional to the object's mass but then F= ma.)

9. Aug 30, 2004

HallsofIvy

[QU0TE= mrjeffy321]"then once it gets ontp the ramp, the gravitational acceleration changes from
being -9.81 m/s^2 down (Y) to a combination of X and Y acceleration."

No. Since you have changed the velocity (up the slope) to
Vx = 10 * cos (60) = 5 m/s
Vy = 10 * sin (60) = 8.7 m/s (the "speed" up the ramp is 10 m/s)
it is clear that x and y here are horizontal and vertical. The acceleration is still
Ax= 0, Ay= -9.81.

10. Aug 30, 2004

Gmaximus

Oh, Duh. I shouldn't try to help people so late at night.

Obviously, g is constant for everything. Funny how stupid i can be.

Sorry Mr.Jeffy, I sincerely apologize for confounding you further.

11. Aug 30, 2004

mrjeffy321

so I shouldnt be trying to change the acceleration into X and Y acceleration, i should leave it as all Y.