A block is on an incline. Derive an equation for the normal force exerted on the block in terms of m (mass of block), g (gravity), and theta(angle of incline)

And you have to start with 2nd-law in the y direction

Are you simply 100% stuck, with no idea at all what to do, or is there some specific point you know you're missing, or what?

my message ain't getting through :(

I can get to F_normy = F_grav, but I'm not sure how to go from there to F_norm

ahah there it is

Can you say anything about the relationship between F_norm and F_norm_y?

@bobble F_normy and F_gravy sound F_unny

F_normx^2 + F_normy^2 = F_norm^2

also there is trig with theta

That's one thing. Anything else?

Mmm, gravy.

But I'm having trouble figuring out how the theta-angle of the slope is relevent to the F_norm triangle

Like, what are the angles in my F_norm triangle?

That's a very good question to ask.

@GarethMcCaughan Hehe. I know Gareth is typing, but it seems like one those things where the person is like "Ah yes" and then moves on

So, tell me more about that triangle. One side is F_norm, right? So what exactly is that side? How long is it and which way does it point?

wait... is the angle between F_norm and F_normy = theta?

You tell me.

F_norm side is perpendicular to incline

(And tell me why, either way.)

That is true.

Okay, if you drop a side from the top of F_norm to the incline, make a side inbetween there and the center of mass, and take F_norm as the third side

and then the angle between the two new sides is 90-theta, because it + the incline angle equals 90

wait, is that the correct triangle?

and the angle between F_norm and incline is 90

(I don't think it is, because it doesn't have F_norm as the hypoteneuse)

(unless I'm visualizing it wrong, which I could be)

yah I changed the triangle

Will make image quick

It looks to me as if you're headed in a direction that will get you there.

(Which I may or may not be, with your very long cryptic clue. I'm pretty sure I know what the answer is and the wordplay for somewhat over half of it, but I'm still working on disentangling the rest.)

the line running SW-NE is the slope?

yes

so then shouldn't the thing perpendicular to that slope be F_norm not F_normy?

oh yeah sorry

and the other two legs of your triangle be horizontal and vertical?

made image too quick

I mean, you may well end up with a triangle substantially resembling that one, but I don't think what you drew can be what you want as it stands.

I think that bottom angle is 90 - theta, because it adds with theta to be perpendicular to flat x-ground plane. And then middle angle is 90, because F_norm is perpendicular to incline. And then top angle is theta

(assume that labeled edge is F_norm)

If the labelled edge is F_norm then this can't be the triangle you need, can it? Aren't you resolving the force into horizontal and vertical components? That triangle doesn't have a horizontal edge, and its hypoteneuse is the vertical edge not the one in the direction of F_norm.

But this is just an extension of the F_norm triangle downwards, and it shares a top angle

so knowing the top angle gives me angle between F_normy and F_norm, and then the other angle

Ah, now that is true.

So now draw me the correct triangle and tell me what you know about its sides and angles.

(elevator music)

Top angle is theta,

Bottom-left angle is 90

bottom-right angle is 90-theta

F_norm is hypotenous, and F_normy is right side and F_normx is bottom

I think you've pretty much got everything you need now. Well done!

cos(theta) = F_normy/F_norm

F_normy = F_norm*cos(theta)

Wait, now I'm getting F_norm = mg/cos(theta). That seems wrong

@bobble Why does F_norm = mg / cos theta seem wrong?

(That's an actual question, not a comment on whether it is or isn't wrong.)

Increasing theta (in 0 < theta < 90) decreases cos(theta). F_norm would increase since mg/cos(theta), and F_norm feels like it should decrease if angle increases

@bobble Let's take the most extreme cases. (1) Suppose the inclined plane isn't actually inclined, it's horizontal, so theta = 0. Then F_norm would be equal to the gravitational force (but of opposite sign). So, first question: does that seem right or wrong? Again, that's an actual question.

cos(0) = 1, so it would be F_norm = mg, which is correct

OK, good.

but when cos(0) = 1, it doesn't matter whether its x or /

other extreme case: theta = 90

cos(90) = 0

Now the other extreme. Suppose the inclined plane is pretty much vertical. Then your formula says that F_norm needs to be very large. Your gut says (above) that on the contrary it needs to be very small. Have a think about the actual forces on the object; is either of those right, and if so which, and why?

normal force should be tiny - block is not being held to inclined plane at all

ah, careful

there are two separate questions here

i mean nonexistant

one is "what will the forces actually be if you put a block on a plank and change the angle of the plank?"

but that isn't the question they've asked you

they've asked: suppose you've got this block on the plank and the forces all balance so that the block stays there rather than sliding down or falling off or whatever; then what do the forces have to be?

and that's not the same question

there's also friction force, which they don't explicity say is ignored

here: moodle.nisdtx.org/pluginfile.php/1368179/mod_resource/content/1/…

(the same workbook, but off the internet)

page 8

@bobble (sorry about delay, was contemplating friction) So, yes, indeed you need to consider friction and it's a little more complicated than what you had before.

You aren't just resolving F_norm into its horizontal and vertical components.

brain is currently hurting

Understood. So, tell me all the forces acting on your object.

F_norm, perpendicular to plane. F_grav, straight down. F_frict, parallel to plane and pointing backwards

Right.

Now, you already worked out the vertical component of F_norm, right? It's F_norm cos theta, upward.

What about the horizontal component of F_norm?

(rightwards)

Why is it rightwards?

or leftwards

sorry i get my directions mixed up

it is in the direction of the avatars in chat, when moving from the chat messages

Correct. (I was going to type "right" but thought better of it.)

I have no idea what F_frict is

Do I need to consider it?

to find F_norm

Now, actually there are two ways we can do this. One way will let you work out the frictional force as well as the normal force, and it's basically straightforward.

I have to start with F_net = ma, remember

The other involves doing one not-quite-so-obvious thing but is less work and will just get you the normal force, which is the one they actually asked you for.

They probably want the second one, but we can do it either way. Or both ways.

Second one then.

Sorry if I'm being difficult

OK. So, earlier you resolved the normal force in the horizontal and vertical directions.

But there are other forces, and other directions you could resolve 'em in.

what other directions? Aren't we only considering x and y?

We can consider whatever directions we want to! Don't you believe in freedom?

um... but the equation they give me is F_net_y = ma_y

so I figured I had to stick with y

Really, it's their starting equation that's confusing me a lot

Oh, I'd forgotten that. In that case we'll do it the other way, which will get us the frictional force too. A little more work, but more profit at the end in exchange.

OK. So you already worked out that the x and y components of F_norm are F_norm sin theta (leftwards) and F_norm cos theta (upwards).

What are the x and y components of F_frict?

F_frict_y = F_norm_y - F_grav

(We are going to need to use F = ma in the x direction as well as in the y direction with this approach. Hope that's OK.)

That's fine, as long as I can fit it

That equation isn't wrong, but it's not the one I had in mind. I was thinking instead of doing the same thing with F_frict that you did with F_norm.

making triangles?

Yup. (You might be able to keep the triangles in your head, but draw them if it helps.)

F_frict_y = F_frict * sin(theta)

F_frict_x = F_frict * cos(theta)

Right.

And, finally and rather trivially, what about the x and y components of the gravitational force?

F_grav_x = 0

F_grav_y = F_grav

Splendid.

F_grav = mg, also

So, now, since (I take it) we are assuming that this thing sitting on the inclined plane isn't accelerating, what equations do you get from these using F = ma in the x- and y- directions?

okay, I'm declaring right to be positive and up to be positive

A reasonable convention.

F_frict_x - F_norm_x = ma = m(0) = 0

OK, so what does that turn into in terms of F_frict and F_norm and a pile of trig functions?

F_norm_y + F_frict_y - F_grav_y = ma = m(0) = 0

(Same question for the y equation, but of course with F_grav in there too.)

F_frict * cos(theta) - F_norm * sin(theta) = 0

F_norm * cos(theta) + F_frict * sin(theta) - F_grav = 0

I have to solve for F_norm

OK, very good. And of course F_grav here = mg.

Right. How are you at solving pairs of simultaneous linear equations?

there's trig in here, which means... not as good

Oh wait if I add them

(I do hope the problem we're solving is actually the one they want you to solve. If e.g. they want you to assume zero friction and not assume no net acceleration of the object, everything becomes simpler...)

Teacher is going to just glance that I did the work, the point here is to make me understand how to derive stuff on inclined planes

Just adding them won't help you, I think. But if you first multiply each equation by something that will make the F_frict terms cancel out?

Well, if the goal is general practice at deriving stuff on inclined planes then we're doing a good thing whether or not it's the intended thing.

How does mutiplying help? I can't divide out, since the term would remain

oh, I just realised something, which means they probably wanted the other approach, but we might as well finish this one first. Won't take much longer.

Happy to move to another room. Hang on a moment.

I'm here

Right. And F_grav = mg.

F_norm * cos(theta) + F_frict * sin(theta) - mg = 0

F_norm * cos(theta) + F_frict * sin(theta) = mg

OK. Now, we aren't really interested in F_frict. The way you get rid of a variable you aren't interested in when you have a pair of simultaneous linear equations is to multiply the two equations by things that give them equal or equal-and-opposite coefficients for the boring variable -- and then you can just subtract or add the equations.

multiply by sin(theta) for x one and cos(theta) for y

Excellent. What do you get when you do that?

doing y - x gives

F_norm * cos(theta) * sin(theta) + F_norm * sin^2(theta) = mg

F_norm * sin(theta) * (cos(theta) + sin(theta)) = mg

wait

something's fishy there

in fact two things are fishy

Check the y equation.

did I subtract wrong?

oh wait its cos^2

cos^2 is one thing. Also check your RHS.

F_norm * cos^2(theta) + F_norm * sin^2(theta) = mg

F_norm * cos^2(theta) + F_norm * sin^2(theta) = mg * cos(theta)

OK!

Now can you work a little bit of trigonometrical magic on the LHS?

F_norm * (cos^2(theta) + sin^2(theta) = mg * cos(theta)

F_norm = mg * cos(theta)

Boom!

YAYAYAYAYAYAYA

thank you so much

it feels like this took too long for me to figure out

And if you do sin and cos instead of cos and sin then you can find what the frictional force is too.

Now.

I'm pretty sure they actually wanted you to do it the other way :-)

but don't worry, it's actually easier.

Okay, I'm listening

So I think that all they were getting at with the F_y = ma_y thing was the observation that the gravitational force equals mg.

yes yes

So, let me say again what I said before: there are a bunch of different forces here and a bunch of different directions you can resolve them in. We already found out what happens when we resolve everything in the x and y directions, and that leads to the answer but requires us to solve some simultaneous equations. Are there any other things to resolve in other directions that might be useful?

You could split gravity into the opposite vectors of F_norm and F_frict

Ah. And what happens then?

And then tilt the plane

and then its's just a normal stationary-cube probem

Do you think that's what I should write?

The plane's already tilted; not sure what you mean by "and then tilt the plane".

un-tilt the plane

so that the "incline" is now flat

No need to un-tilt anything.

You could also declare x and y in relation to the incline plane's tilt

You can resolve any vector into its components in any two perpendicular directions -- they don't need to be horizontal and vertical.

And if you draw one of those little triangles, it will end up looking ... rather ra lot like the one you started with.

So I would need to find F_grav's components

Right.

Which are equal in magnitude to F_norm and F_frict

And that diagram will be right not wrong, this time, because its hypoteneuse is now F_grav, which is the thing you're resolving.

Yup, components are F_norm and F_frict

And the angle is still theta for all the same reasons as before.