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sutabi
10-04-2003, 08:09 AM
I am wanting a constant acceration no matter incline

The incline is 15 degree but I dont know it 15 degrees, the gravity is earth gravity and the mass is 1. The time interval = .1

I just take the current position - last stored position to find the x and y displacment

so the resulting force or net force = mass*gravity*(y/sqrt(x*x+y*y)

Note: (y/sqrt(x*x+y*y) = cos(theta)

so force = .855 and veloctiy (x axis) = .29

Now lets say I want velocity to be a constant 1.00

So displacment(x axis) = v/t = 1/.1 = 10
and I can't calculated acceration until the postions are stored
so A = final velocity - initial velocity/time interval

NOW! here where I am lost if I can figure the force needed for 15 degress and the velocity is .25 how do I figure out the force need to keep velocity at 1.00 with x degrees

ajk48n
10-16-2003, 01:38 AM
I'm not quite sure if this helps or not, but you said:

so the resulting force or net force = mass*gravity*(y/sqrt(x*x+y*y)

Note: (y/sqrt(x*x+y*y) = cos(theta)

Wouldn't this also mean that

force = mass*gravity*cos(theta)

If it did, then coudn't you find the force by just plugging the angle of your ramp into the equation?

Also, in this:

So displacment(x axis) = v/t = 1/.1 = 10
and I can't calculated acceration until the postions are stored
so A = final velocity - initial velocity/time interval

You say you can't find A until you have the positions, but can't you assume that one position will be (0,0), and then find the other position based on veloctiy being 1, and you angle being x.

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