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There is no tangens involved in the problem.
There's a useful check you can do on whether you have chosen correctly between sine and cosine. What if we change the angle to 0, so that the ramp is horizontal? If it's sine, that would make the vertical component of the normal force zero. Does that seem right?Shaku said:In that case it would be:
Fny = FnSin(30) (Since slope is Fn, angle is 30, and we're looking for the opposite)
haruspex said:There's a useful check you can do on whether you have chosen correctly between sine and cosine. What if we change the angle to 0, so that the ramp is horizontal? If it's sine, that would make the vertical component of the normal force zero. Does that seem right?
haruspex said:Good. Now have another go at the horizontal and vertical components of the two forces.
Yes! (Well, almost: should be mg, not g.)Shaku said:- what is the component of Fn in the vertical direction?
FnCos(30)
- what is the component of Fn in the horizontal direction?
FnSin(30)
- what is the component of Fg in the vertical direction?
-g
- what is the component of Fg in the horizontal direction?
0
haruspex said:Yes! (Well, almost: should be mg, not g.)
Next, we have to relate these to acceleration. Suppose the acceleration down the ramp (i.e. parallel to the ramp) is a. What are the vertical and horizontal components of that acceleration? What equations can you then write down relating these to the four force components above?
The normal force and the gravitational force are in almost opposite directions. Which direction (up or down) are you taking as positive? Also, you've dropped the m from mg again.Shaku said:ay = (FnCos(30)+g)/m
haruspex said:The normal force and the gravitational force are in almost opposite directions. Which direction (up or down) are you taking as positive? Also, you've dropped the m from mg again.
Yes.Shaku said:ay = (FnCos(30)/m) - g
ax = FnSin(30)/m
No, that's not how you add vectors.thus the acceleration is:
a = ((FnCos(30)/m) - g) + (FnSin(30)/m))
Yes, but as I said, the next step is think about the ratio between the horizontal and vertical accelerations. Or equivalently, if a is the nett acceleration down a ramp at 30 degrees, what would its horizontal and vertical components be?Shaku said:So, it would be:
a = [itex]\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}[/itex]
haruspex said:Yes, but as I said, the next step is think about the ratio between the horizontal and vertical accelerations. Or equivalently, if a is the nett acceleration down a ramp at 30 degrees, what would its horizontal and vertical components be?
Well, yes, but all I was looking for was ay=-a sin(30), ax = a cos(30).Shaku said:ay = Sin(30)*[itex]\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}[/itex]
ax = Cos(30)*[itex]\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}[/itex]
Oops, hahaha! I noticied when I first wrote it down I had them backwards, so I just switched the first letters without realizing what I did!haruspex said:I assume you meant: ...
haruspex said:So what is ay/ax as a function of theta?
What equation does that give you if you equate it to ay/ax from these equations:
ay = (FnCos(30)/m) - g
ax = FnSin(30)/m
If ay=-a sin(30) and ax = a cos(30), what is ay/ax?Shaku said:I'm not sure what you're mean by "as a function of theta"...
It will allow you to work out Fn/m. You don't need to know the individual values.So you're saying to set each of the two ay/ax expressions equal to each other and then solving for a variable? Wouldn't that still leave me with either mass or Fn that didn't cancel out?
haruspex said:If ay=-a sin(30) and ax = a cos(30), what is ay/ax?
It will allow you to work out Fn/m. You don't need to know the individual values.
No, that would just give 30=30.Shaku said:So you're saying:
30 = Tan-1(-aSin(30)/aCos(30)) and then somehow solve for a...? (Since to find an angle given two sides, you take the inverse tan of the two sides).