# Exactly Not a Homework Question, but something that I can't solve

1. Jan 10, 2005

### sastory

The Question is This:

There is a slope shaped y=e^x. A point mass of 1kg is let go from the position x=10 so that it slides to position x=0. Assume no friction.
Find N(t), a function of normal force's magnitude in respect to time.

I have tried solving it using various ways, but I always get shot down by calculus. Any help?

2. Jan 10, 2005

### so-crates

What two forces are acting on the body as it slides down the slope? What must the direction of the total force be?

3. Jan 10, 2005

### sastory

obviously I know that they are normal force and gravity - I have set y-axis as my gravity since normal force's direction keeps changing.

4. Jan 10, 2005

### vincentchan

hints:
the slope is e^x.....why?

edit:
one more hints:
$$\tan{\theta} = slope$$

Last edited: Jan 10, 2005
5. Jan 10, 2005

### HallsofIvy

Staff Emeritus
No, the two forces are NOT "normal force" and gravity. The only force acting is gravity. You can reduce that to components that are normal to the curve and tangent to the curve. The normal component only holds the partical on the curve and does not affect its speed. The component tangent to the curve is what accelerates it along the curve.

6. Jan 10, 2005

### sastory

I got the slope part. I even got the equation for Nx(x)... but the problem is x in this function is in fact x(t) and to get that I have to integrate Nx(x)/m. I don't know how.

I tried to get component tangent to the curve, but this is hard for me since the tangent changes over t... I really don't know how.

7. Jan 10, 2005

### vincentchan

you need x(t), once you have both x(t) and N(x), N(t)=N(x(t))-->> plug it in and see the answer

8. Jan 10, 2005

### sastory

but then how will I get x(t)? don't I need a force equation to do that (which also refers to x(t))? I tried it doing it by using delta KE = delta PE, but it doesn't seem to do that.

9. Jan 10, 2005

### sastory

This is what I have for Nx(x(t)):

Nx(x(t)) = mgsin(e^x(t))cos(e^x(t))
ax(x(t)) = gsin(e^x(t))cos(e^x(t))

10. Jan 10, 2005

### vincentchan

obviously, what you did is WRONG, you have
$$\tan{\theta}=e^x$$
$$\sin{\theta}=e^x/ \sqrt{1+e^(2x)}$$
$$\cos{\theta}=1/ \sqrt{1+e^(2x)}$$

draw a triangle and you the P**** therom, you will see the result above.... and do the calculation yourself, some integration is neccissery for x(t), I am not sure am i allowed to post the answer here, I have been warned by some administrater this morning......

11. Jan 11, 2005

### Galileo

The normal force does act on the particle. As you said: it's the normal force that keeps the ball on track. If only gravity is acting on the ball it would fall straight down.
The normal force does not affect its speed, since the direction of the force is alsways perpendicular to the motion of the ball.