Sliding Ramp and Freefall problem

[lep11]

Do you know what the answer should be?

 

[lep11]

Btw, neglecting air drag should have been mentioned in problem statement.

 

[Shaku]

The diagram is right now but those equations do not spring from it at all. What happened to the normal force? How does g get a horizontal component?
Let's try it again, one step at a time:
- what is the component of Fn in the vertical direction?
- what is the component of Fn in the horizontal direction?
- what is the component of Fg in the vertical direction?
- what is the component of Fg in the horizontal direction?

- what is the component of Fn in the vertical direction?
9.8 (Since Fn is opposite of g).
- what is the component of Fn in the horizontal direction?
I'm not sure... since you said this wasn't gCos(30), but that's the only way I see that I can get this force.
- what is the component of Fg in the vertical direction?
-9.8 (gravity)
- what is the component of Fg in the horizontal direction?
0

Edit: At the bottom it says "in the downward direction"
Edit2: Just realized my math was wrong for F_nety.
Should be:
9.8Sin(30) - 9.8 = 4.9.
Edit3: Also just realized this doesn't work either... It just = 9.8 again... Sigh. I need help ><

Do you know what the answer should be?

Sadly, no.

Btw, neglecting air drag should have been mentioned in problem statement.

This problem was given to us before any force concepts were introduced, so it was assumed that air drag is neglected.

 

[haruspex]

- what is the component of Fn in the vertical direction?
9.8 (Since Fn is opposite of g).

The acceleration will have a vertical component, so you can't assume the forces balance. At this stage, I'm not asking you to equate forces. My question is much simpler: if the normal force is Fn, and the slope is 30 degrees, what is the component of Fn in the vertical direction?
Similarly my other questions.

 

[Shaku]

My question is much simpler: if the normal force is Fn, and the slope is 30 degrees, what is the component of Fn in the vertical direction?
Similarly my other questions.

In that case it would be:
Fn_y = FnSin(30) (Since slope is Fn, angle is 30, and we're looking for the opposite)I was talking to a friend earlier today, and he said that the force at the end of the ramp would be gTan(30), and then you'd use that to somehow work out the problem. I'm still completely lost though...

If he's correct, then Fn_y = gTan30*Sin(30)... But that seems like a really small force (~0.28867 Newtons), so I doubt that's correct.

 

[mfb]

There is no tangens involved in the problem.

 

[haruspex]

In that case it would be:
Fn_y = FnSin(30) (Since slope is Fn, angle is 30, and we're looking for the opposite)

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]

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?

So, in my diagram (I know I drew it wrong, just imagine the red triangle is touching the slope. The slope is the X axis, and perpendicular is the Y axis), you're saying that the vertical component Fn_y is FnCos(30)? SohCahToa - Shouldn't it be Sin since we have the hypotenuse and are looking for the opposite angle (which is the vertical component of Fn)?

Why is it that it's FnCos(30) and not FnSin(30)? I may just be picturing this wrong, but if we use FnCos(30), wouldn't we be finding the horizontal component instead of the vertical component?

 

[haruspex]

It's called the normal force because it is normal to (i.e. perpendicular to) the surface. You're drawing it at an angle of 60 degrees to the surface.

 

[Shaku]

It's called the normal force because it is normal to (i.e. perpendicular to) the surface. You're drawing it at an angle of 60 degrees to the surface.

Ohhhhh, that clarifies things a lot!

It would be like this:

What do I do next?

 

[haruspex]

Good. Now have another go at the horizontal and vertical components of the two forces.

 

[Shaku]

Good. Now have another go at the horizontal and vertical components of the two forces.

- 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?
mg
- what is the component of Fg in the horizontal direction?
0

 

[haruspex]

- 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

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?

 

[Shaku]

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?

Ah, you're right; I forgot about the mass there.

a = F_net/m

So vertical acceleration is:
a_y = (FnCos(30)+g)/m

and horizontal acceleration is:
a_x = FnSin(30)/m

thus the acceleration is:
a = ((FnCos(30)+g)/m) + (FnSin(30)/m))

 

[haruspex]

a_y = (FnCos(30)+g)/m

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]

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.

g = -9.8 as in down is negative
(I changed it to -g below for clarification)

and oops!

a = Fnet/m

So vertical acceleration is:
ay = (FnCos(30)-mg)/m
ay = (FnCos(30)/m) - g

and horizontal acceleration is:
ax = FnSin(30)/m

thus the acceleration is:
[STRIKE]a = ((FnCos(30)/m) - g) + (FnSin(30)/m))[/STRIKE]

 

[haruspex]

ay = (FnCos(30)/m) - g
ax = FnSin(30)/m

Yes.

thus the acceleration is:
a = ((FnCos(30)/m) - g) + (FnSin(30)/m))

No, that's not how you add vectors.
Next, we need to make use of the fact that we know which direction the overall acceleration will be in: down the ramp. This tells us the ratio of the two acceleration components. Can you see how to do that?

 

[Shaku]

So, it would be:
a = \sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}

So how I do I find the acceleration in terms of an actual number once I have this?

 

[haruspex]

So, it would be:
a = \sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}

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]

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?

ay = Sin(30)*\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}
ax = Cos(30)*\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}

 

[haruspex]

I assume you meant:

ay = Sin(30)*\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}
ax = Cos(30)*\sqrt{(((FnCos(30)/m) - g)^2 + (FnSin(30)/m))^2)}

Well, yes, but all I was looking for was a_y=-a sin(30), a_x = a cos(30).
So what is a_y/a_x as a function of theta?
What equation does that give you if you equate it to a_y/a_x from these equations:
ay = (FnCos(30)/m) - g
ax = FnSin(30)/m

 

[Shaku]

I assume you meant: ...

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!

So what is a_y/a_x as a function of theta?
What equation does that give you if you equate it to a_y/a_x from these equations:
ay = (FnCos(30)/m) - g
ax = FnSin(30)/m

I'm not sure what you're mean by "as a function of theta"...

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]

I'm not sure what you're mean by "as a function of theta"...

If ay=-a sin(30) and ax = a cos(30), what is ay/ax?

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?

It will allow you to work out Fn/m. You don't need to know the individual values.

 

[Shaku]

If ay=-a sin(30) and ax = a cos(30), what is ay/ax?

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).

It will allow you to work out Fn/m. You don't need to know the individual values.

Ah, okay.

 

[haruspex]

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).

No, that would just give 30=30.
We have:
ay/ax = -aSin(30)/aCos(30) = -tan(30)
ay = (FnCos(30)/m) - g
ax = FnSin(30)/m
Combine those three equations, eliminating ay and ax. That will give you one equation that will tell you the value of Fn/m.