# Mechanics question not even that advanced

1. Feb 20, 2006

### schattenjaeger

So if you have a block sitting on top of a frictionless slide(it's a ramp, so it goes down like a slide, levels out at the bottom at point A, then curves back up and ends at point B.)the height at the top of the slide is 3.91m, and the mass of the block is 1.62kg, the angle at point B is 45 degrees from the horizontal(err, in case that wasn't clear, at point B the block goes flying off at 45 degrees)and the height of point B is not given

so anyhoo, it wants to know the force exerted on the block by the track at point A(where it levels out) wouldn't that just be mg? That seems too simple though. Same thing at point B but with some trig fiddling to get like cos(45)*mg. I guess it COULD be that simple, there's two more parts to the problem that are giving me LESS problems just because I'm not second guessing myself, but I dunno.

Edit: I guess that doesn't really make sense though. In that case the force is constant, but obviously as you go faster and pull up like that, you're applying a greater force down(and hence the track is pushing up harder)

Last edited: Feb 20, 2006
2. Feb 20, 2006

### schattenjaeger

Well I found the solution online, but I wanna call BS! I figured it would involve centripetal acceleration in some way, but it gives me an R and angle, but by the time it reaches point B it's lost that circular motion clearly in the picture, and it's not stated in the problem. Or something, I just suck I guess

3. Feb 20, 2006

### steelphantom

It really is that simple. At point A, the force exerted is mg, and at point B it would be m*sin(45). And you're right; this isn't exactly advanced Physics. :tongue:

Edit: I'm confused now. Could you get a diagram of what this situation is supposed to look like?

4. Feb 20, 2006

### schattenjaeger

Yah, like I said the problem kinda fudges on its explanation. Later in the problem I need to do calculations at point B using that same R that's given, but in the picture they're clearly not the same distance from the center of the "circle" but whatever

5. Feb 20, 2006

### schattenjaeger

http://img152.imageshack.us/my.php?image=untitled3ze.png
exagerated features 'cuz I suck at drawing, especially on comps
A's at the bottom, B's the end point, it makes perfect sense if you assume R is the same length as the line going to B, and then it's fine, but you can't really assume that, can you? Obviously the problem does, I'm just saying. The answer, fyi, is Normal force=mg(1+2h/R)