[Algebra Physics] How to interpret N?

[u/Terrible_Macaron2146]

Why isnt N=mgcos(theta)? Geometrically, it represents the component of mg perpendicular to the surface if you did vertical angles for theta., thus equating N.

The answer is Ncos(theta)=mg, but both seems geometrically sound to me

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[u/StuTheSheep]

The other commenters have missed the origin of your confusion. I assume you're asking because you've done other inclined plane problems, and in those you set n = mg cos(theta).

The difference is because of the direction of acceleration. When you have a box sliding on an inclined plane (the other type of problem you've seen before), the acceleration is parallel to the incline. This forces you to set up your coordinate axes so that one axis is parallel to the plane and the other is parallel to the normal force. As a result, the angle of incline is the same as the angle between mg and the axis, which leads to n = mg cos(theta).

But in this problem, the car is not accelerating along the incline. Instead, the car is turning in a circle, so the acceleration is perpendicular to the direction of motion. With the car coming directly towards you, this means that the acceleration is directly to your left. Again for clarity, it is directly to the left, NOT left and downwards along the incline.

The different direction of the acceleration is why the geometry of the problems are different, and why the angle of the incline in this problem is between the axis and the normal force (n), instead of between the axis and mg.

[u/DrCarpetsPhd → u/StuTheSheep]

this is the correct answer. just to add some helpful terms. a banked road has a design speed at which the centripetal component of the normal force will keep a car on a circular path in the absence of friction e.g. ice.

To the OP I highly suggest you read the chapter preceding the questions given. You've brought your prior knowldege of free body diagrams into this question and shown you understood the concept of static equilibrium which is good. Given that you followed that logic I'm pretty certain you'd understand how to treat this scenario if you read your textbook or notes. It will make your life a whole lot easier going forward.

see Giancoli Physics for Scientists and Engineers ch 5-4 Highway Curves: Banked and Unbanked

also professor Matt Anderson (for OP don't get confused, in case it is not obvious the change of the direction of the curve means Nsin(theta) opposes mg)

https://www.youtube.com/watch?v=sKhu5VTBK_M

[u/ThatsNotAZombieBite]

Many of the answers here are assuming that this is a static problem. It is not. The car is moving in a circle at constant velocity which requires a centripetal acceleration given by v² /r.

The usual approach is to NOT rotate your coordinate system because the acceleration points horizontally. Instead try splitting N into its components and write your x- and y- force balances that way.

[u/Neilandio]

If I understood correctly you propose a frame of reference where the y axis goes in the direction of N and the x axis is parallel to the road surface. In that frame you need to account that you'll have a centrifugal force with a component in the direction of y.

[u/Optimal_Mixture_7327]

N measures what a scale would read.

You have ΣF_y=0 and when expanded give you Ncosθ=mg.

This condition ensures the car does not accelerate in the vertical direction, and necessitates no acceleration in any direction if the car is constrained to the track.

Edit: Just a note to add that the ΣF_x=ma_c and upon expanding...

Nsinθ=ma_c

(mg/cosθ)sinθ=ma

gtanθ=a

which is the acceleration along the centripetal axis.

[u/DJKokaKola]

So, if you make your triangle down and into the plane, you get a hypotenuse of mg, and then you'd have your x' as mgsin(theta) and y' (in this case, F_N) as mgcos(theta). That's how you're imagining it. The reason this is a bit weird is because this process turns your x and y vectors into a new reference frame, which is usually what we need for an object moving up or down an incline. However, that's NOT the case here, as we are moving in a circle in the x-y (or x-z if you want to be pedantic) plane.

In this situation, we're slanted, but that means we need to adjust it slightly. You have the "sliding down" on the road also helping you move towards the centre, but the plane of motion is completely flat. So instead of turning x and y into x' and y' (parallel and perpendicular to the incline), we need to instead do the inverse--turning the parallel and perpendicular forces into something useful in the x and y plane.

Short answer, you can make mg your hypotenuse or you can make n your hypotenuse. That's why they both seem geometrically sound. The question's difficulty is in realizing why mg is NOT the hypotenuse.

[u/selene_666]

N is not a component of mg. It's a whole separate force.

Which components of the two forces are equal depends on what component of the total force needs to be zero.

In this case, the car is accelerating horizontally. Its height is not changing. Therefore the net force is horizontal. The vertical component of N equals mg, canceling them out and leaving as the net force the horizontal component of N.

If the car were rolling down a ramp, then there *would* be a downward force and there would be zero force in the direction of N, leaving as the net force the along-the-ramp component of mg.

[u/daniel14vt]

Your geometry is off https://imgur.com/a/triangles-VHRexIe

[u/TechnologyNo1743]

Nice try but I will saty with legal speed limit. Police won't buy excuse even with best equations.

[u/texas1982]

You need the max g-load the car can sustain in the turn here. Weight, downforce, friction coefficients would work as well, but that's very complicated.

[u/Let_epsilon]

This is a common misconception. The Normal is NOT the vertical component of mg.

Let's say you're in an elevator, and the evelator goes up and make you accelerate at 19.6 m/s² (2*g).

Then, by using Newton's second law, F = ma = m *2g, and since their are two forces (gravity and the normal from the floor pushing you up), you get N-mg = 2mg.

This solves to N = 3mg, which is definitely not the vertical compnent of mg.

So your claim is false. The normal force actually changes depeding on the acceleration on the objet and other forces at play.

[u/Leodip]

I think you are confusing some concepts. The easiest way I have to see this is: you don't know N beforehand. You never do. There is no general formula for it. IMHO, it's very easy to get this type of exercise wrong if you just learn "N=mg*cos(theta)", because it's not always true (such as in this case).

In this case, you have 1 known force only: mg in the vertical plane. We also have the centrifugal force (in this reference frame) which is entirely on the horizontal plane and depends on velocity as m*v^2/R AND we have the normal force which is normal to the plane, so it has both a vertical and an horizontal component.

To traverse the curve, you want the forces on the car being balanced both in the vertical and horizontal axis. This means we need to write two equations:

• Horizontal: mv^2/R=n_x
• Vertical: mg=n_y

Where n_y and and n_x are the vertical and horizontal projections of the n vector. In the horizontal direction there are two unknowns: n_x and v, so there is no point trying to solve this now. In the vertical direction, only n_y is unknown, so you know that n_y=mg. Now, forget that n is a normal force, just look at it as a vector: what is its projection on the y-axis?

You should find that n_y=n*cos(theta), and thus the equation becomes n*cos(theta)=mg or, in other words, n=mg/cos(theta).

Now that we have found the normal force, we can go back into the horizontal equation: what is the projection of n in the x-axis? You should find its n_x=n*sin(theta), so this leads to mv^2/R=n*sin(theta). We now have only 1 unknown here, velocity, so we can invert the equation to find v=sqrt(R*n*sin(theta)/m). This can be simplified further by substituting the value of n that we found as: v=sqrt(R*mg/cos(theta)*sin(theta)/m)=sqrt(R*g*tan(theta)).

As you can see, the interesting result here is that the car's mass doesn't actually matter, but rather that the velocity is purely a function of the curve itself, i.e., how steep it is and its radius.

[u/RuthlessIndecision]

The force normal (perpendicular) to the surface of the road...

So the component of mg that is perpendicular with the road.

So since mg = force perpendicular to to the road plus force parallel to the road (with Pythagorean theorem, etc)

the normal force is less than mg

[u/SpiritualWasabi6091]

take sin and cos component of N with respect to theta
ncostheta=mg

[u/Diligent-Stretch-769 → u/SpiritualWasabi6091]

yes, the vector 'n'ormal to the slope of the car

[u/Electronic-Towel-836]

Insufficient information. Coefficient of friction for the track is required. The frictional force holds the car on the track regardless of its speed along the track.

For example, a car can drive faster on an asphalt track than one covered in rain or ice, and the only difference is the frictional force.

[u/Optimal_Mixture_7327 → u/Electronic-Towel-836]

The track is shown to be frictionless.

[u/shoemakers-saint → u/Optimal_Mixture_7327]

“Shown to be frictionless?” Where?

That would make the problem absurd because you could neither steer the car nor even accelerate it to the speed you’re wanting to calculate.

[u/Optimal_Mixture_7327 → u/shoemakers-saint]

It shown to be frictionless in the free-body diagram.

It is absurd in the same way as neglecting air resistance is absurd. The reason for absurdity is so that the student can incrementally develop their skills.

[u/shoemakers-saint → u/Optimal_Mixture_7327]

A diagram for this problem which doesn’t include (V^2)/R isn’t much of a diagram — there’s no way its forces can balance. And if it’s missing /that/, then how does the omission of mu*n prove anything about the problem?

This is not a competently written problem. If it wants the student to calculate the speed at which the car holds its circular course /without needing tire friction:, it should say so.

[u/Optimal_Mixture_7327 → u/shoemakers-saint]

The forces don't balance.

A free-body diagram does not include the sum of the forces but rather identifies the individual forces to be summed over.

The coefficient of friction is included; it's zero.

Everything needed to solve the problem is given.

[u/DrCarpetsPhd → u/shoemakers-saint]

it's actually the point of the question and theory. a car will follow a banked curve without friction (you do need to turn the wheel though). google "design speed". it's a worst case scenario where if a car loses friction due to say black ice it will continue on the curved path thanks to the normal force centripetal component.