So why is normal force at inclined plane is $mg \cos a$ and friction is $mg \sin a$? I mean, why not vice-versa or why not some other ratio? Where does it come from? I can see that the sum squares of friction and normal force under the root should be equal to $mg$ and the angle at which the plane is inclined has something to do with it.

Does it come from some other part of physics or was it just deduced experimentally? I mean, we could slide a sample block from an inclined plane to see what acceleration it has each time we change the angle.

UPD: Thank you all guys! What I was ultimately asking just seems to be a matter of philosophy. These all formulas are beautiful and seem to be correct. But they are abstractions. And the confidence that this is true ultimately comes from experiments. Since I am very new to physics, I just wanted to know if my assumptions were true.

Normal force at inclined plane


4 Answers 4


In vector notation there is only one equation and no ambiguity. The block is in equilibrium so the net force acting on it must be zero. There are three forces acting on the block - its weight $\vec W$, the normal force $\vec N$ and friction $\vec F$. So we have

$\vec W + \vec N + \vec F = 0$

Since $\vec N$ and $\vec F$ are orthogonal to one another (at right angles) it is convenient to resolve the three vectors into components along an $x$ axis that is parallel to $\vec F$ and a $y$ axis that is parallel to $\vec N$. In component form we have

$\vec F = (F, 0) \\ \vec N = (0, N) \\ \vec W = (-W \sin (a), -W \cos (a))$

The last line comes from the fact that $\vec W$ is at an angle $a$ to the negative $y$ axis - it is simply the rule for how you turn a vector into components. Adding the three vectors and equating $x$ and $y$ components to zero gives two equations:

$F - W \sin (a) =0 \\ \Rightarrow F=W\sin (a)=mg\sin (a) \\ N - W \cos (a) =0 \\ \Rightarrow N=W\cos (a)=mg\cos (a)$

There is nothing special about this choice of $x$ and $y$ axes. We could instead choose axes that are horizontal and vertical. Relative to horizontal and vertical axes the component forms of the vectors are:

$\vec F = (F\cos (a), F\sin (a)) \\ \vec N = (-N\sin (a), N\cos (a)) \\ \vec W = (0, -W)$

If you add the three vectors and equate each component to zero you get two different equations. But if you then solve to find $F$ and $N$ in terms of $W$ you get exactly the same result as before - there are just a couple of extra steps in the algebra.


First, the normal force

  • A normal force is the name we give to the perpendicular force (perpendicular to the surface). It is not equal to $mg\cos(\alpha)$ in general. Maybe it is in this specific scenario but that is just a coincidence.

In your specific case where the object is stationary (I am assuming that it isn't accelerating), if you want to calculate the normal force then you will set up Newton's 1st law in this perpendicular direction. Since only the normal force $n$ and one component of the weight $w$ acts along this direction, then we get:

$$\sum F=0 \quad\Leftrightarrow\quad n-w_\perp=0 \quad\Leftrightarrow\quad n=w_\perp.$$

Now you just need to find the perpendicular component of the weight. That turns out to include the cosine of the angle due to trigonometry: $w_\perp=mg\cos(\alpha)$. If your question is why this cosine appears here, then let me know in the comments and I'll adjust the answer. So, now we know that


and this is not a feature of the normal force. This is only an expression that holds true in your specific scenario. If other forces acted or if acceleration was present, then this expression would look very different.

Now on to friction

  • A friction appears due to the way microscopic interlocking and asperity merging takes place when two surfaces are in contact. Some of these microscopic effects do exerts forces that aren't parallel to the surface. But all such microscopic forces' perpendicular components are typically always just included in the normal force. Then, by definition, a friction force is always fully parallel to the surface.

We do have a formula for kinetic friction, when the object is in motion. But in your case it seems the object is stationary. Then we are dealing with static friction for which we do not have an exact formula! Just like with the normal force, static friction is not in general equal to $mg\sin(\alpha)$.

Use the same argument as above: set up Newton's first law, this time in the parallel direction. You will then see that static friction (which isn't drawn on your drawing - it should be pointing upwards along the surface). is equal to the parallel component of the weight:


Then the question is again what value this parallel weight component has. It turns out to be $w_\parallel=mg\sin(\alpha)$, which is seen from trigonometric considerations again. Again, if it is not clear why the sine appears in this expression and how the trigonometry works, then let me know in the comments and I'll expand the answer.


This question boils down to basic trigonometry and vectors in 2D space. Given a vector, a quantity with a direction and magnitude. Let's consider this magnitude be its length and its direction is given by the arrows that are used here. Now for a vector with a given length, and making an angle of $\theta$ with the x-axis, the projection of it along the x-axis and y-axis will be mgcos$\theta$ and mgsin$\theta$ respectively. We do the same for the force mg to obtain its components along the two directions. This way the square sum of the components adds up to mg, consistent with what happens in geometry. Now why we use geometry and trigonometry here, can it be experimentally proven to be right, and if this is the exact ratio it splits with may be venturing into Philosophy I guess.

  • $\begingroup$ Yes! This is exactly what I was asking about. I can see that this is a mathematically beautiful thing that a force in our case can be split into two different forces. But what I was thinking of was why did we just decide that this splitting is correct, even it's mathematically beautiful and seemingly correct? I was guessing there must have been experiments before stating that this is true. And yes, you answered that this is rather a matter of Philosophy as to why experiments prove this true. We take it for granted. I mean, this thing is not deduced from something else, it's just the way it is $\endgroup$
    – SkyFlame
    Aug 11, 2021 at 12:51
  • $\begingroup$ Maybe it all was silly what I was asking about. But it's just important to me. I couldn't wrap my head around it. Now I understand it better that we should take many things for granted. That this is just how the world works. And, of course, they are proven experimentally. I will wait some time, just in case somebody has another idea of what it's all about (which I doubt). Then, I will nominate your answer as a solution. Thank you! $\endgroup$
    – SkyFlame
    Aug 11, 2021 at 12:53
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    $\begingroup$ Glad my answer was able to give you some clarity. This is just my take on it. There may be a better description that can more convincingly explain this. You are right, many things we take as it is, and the more deeper we think about it, we start questioning the description of reality as we know it. Eventually, we just take it for granted as many experiments, theories that are based on these fundamentals have shown convincing results. When there comes an instance where we cannot explain what we observe, we try and find a better model that fits. $\endgroup$ Aug 11, 2021 at 17:34

In general, forces can be added or decomposed into sums. So if there is a mass point $m$ and two forces $\vec F_1$ and $\vec F_2$ act on it, the total force acting on $m$ is $\vec F = \vec F_1 + \vec F_2$ and it will be accelerated with $a = \vec F / m$. Likewise, if you have a force $\vec F$, you can always take an arbitrary force $\vec F_1$ and define $\vec F_2 = \vec F - \vec F_1$, so that again $\vec F = \vec F_1 + \vec F_2$. Then you can pretend that $\vec F_1$ and $\vec F_2$ are acting on $m$ instead of $\vec F$ and the behaviour of the system will be the same.

For the inclined plane, the gravitational force $mg$ is just not very convenient for describing the system. It is much easier to use a force parallel and another one perpendicular to the plane, because the first one will accelerate masses and the second can be used for determining the amount of friction which inhibits their motion. So in the symbols used before, the gravitational force would be $\vec F$ and it should be decomposed into $\vec F_1$ parallel to the plane and some other force $\vec F_2$. Because of the way (Euclidean) geometry works, $\vec F_2$ will automatically be perpendicular to that same plane. Since $\vec F_1$ and $\vec F_2$ are thus perpendicular to each other, $\vec F,\vec F_1,\vec F_2$ form a right triangle and the ratio of the lengths of the forces can be calculated as sine and cosine.

  • $\begingroup$ But why is mg decomposed using trigonometry (namely, Pythagorean theorem)? Why is there no other ratio? Or does it come from the fact that the normal force should always be perpendicular to the plane? Then, we should find the "gravity" force vector that opposes the normal force. And the only pair of vectors (sum of which gives mg vector) are those vectors that are catheti to the mg vector. All other pairs will not give us a vector (one from the pair) which is perpendicular to the plane. Is this right? $\endgroup$
    – SkyFlame
    Aug 11, 2021 at 11:08
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    $\begingroup$ @SkyFlame You can of course decompose differently. Only, like stated in my answer, the decomposition into the perpendicular and parallel vector is very handy, so most people do it this way. And there is, as you said, only one possible choice for the length of the catheti once the inclination angle of the plane is known and one of them should be perpendicular to that plane. $\endgroup$
    – sim0
    Aug 11, 2021 at 11:16
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    $\begingroup$ @SkyFlame You could decompose differently, but then you would not be talking about the normal and friction forces. Those are defined to be perpendicular and along the interface plane. (Normal is a mathematical term that means perpendicular.) $\endgroup$
    – garyp
    Aug 11, 2021 at 11:44

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