Calculating the component of the normal force (anti-)parallel to the gravitational force and setting them equal, I get \begin{equation} N \cos \theta = m\, g. \end{equation} On the other hand, calculating the component of the gravitational force (anti-)parallel to the normal force and setting them equal, I get \begin{equation} m\, g\, \cos \theta = N. \end{equation} Where have I gone wrong?

My drawings of the calculation

It's being assumed as a duplicate of an already answered question which I think is more about specifics. It addresses two separate diagrams with two situations whereas mine is more general and has different situation than the one in the question. It's like we look alike but fingerprints are different and fingerprints are what matter.

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    $\begingroup$ Is the block at rest? $\endgroup$ Commented Jan 23, 2018 at 14:43
  • $\begingroup$ Whatever you wanna assume,I’m just pointing this out though a bit of context might be needed,well then let us assume it’s going down the direction parallel to the incline and all the surfaces exert friction(frictionless will work tooo) $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 15:05
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    $\begingroup$ Possible duplicate of Why is $R\cos{a} = mg$ in circular motion compared and not $R = mg\cos{a}$? $\endgroup$ Commented Jan 23, 2018 at 15:16
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    $\begingroup$ Two lessons here. One is the physics, but the other is about communications and technical writing: what you did wrong in writing this post was not specifying the situation that the diagram corresponds to. Thus some answers assume a "sliding down a frictionless ramp" scenario (because that is a common problem that involves a FBD like this), and you only got answers assuming "going around a banked turn" after you specified that you were interested in that case. It's good to learn this lesson early, so now you're ahead of the game. $\endgroup$ Commented Jan 23, 2018 at 20:30
  • $\begingroup$ @dmckee The problem described here is a block sliding down a frictionless ramp. Not because it is a common problem, but because that is what is shown in the FBD. If there is friction, it is shown in the FBD. If the block has a forward velocity and is rotating a round an axis, it would be mentioned in the question. $\endgroup$
    – Orbit
    Commented Jan 17, 2019 at 22:10

8 Answers 8


For any given case there are some obvious details that must not be contradicted. A FBD provides a more mathematical approach towards the question but how we form our equations is based on these critical details.

A correct FBD satisfies all these details (or observations).

For the given case, the observations are:

  1. The wedge is stationary,
  2. The block never looses contact with the wedge, and
  3. The block accelerates along the wedge.

When you resolve the normal force vertically and horizontally, you come to two conclusions.

Conclusion $(1)$, Vertical Equilibrium.


This equation indicates that the block is in vertical equilibrium, but this contradicts with observation $(3)$. The block must move vertically downwards.

enter image description here

Conclusion (2), Horizontal acceleration.


In this case, we know that the block never leaves the wedge but if the block has some net horizontal acceleration (as shown in the diagram), it will definitely leave the wedge. This contradicts observation $(2)$.

So, the equations that we come up with using this FBD must be wrong because they don't satisfy our observations. However, it does not mean that the FBD is wrong but the equations we come up with the FBD are wrong.

$$N\cos{\theta} \neq mg$$

When you resolve the weight of the block along and perpendicular the wedge, the resultant FBD satisfies all our observations. The two correct equations are,

Perpendicular to the surface: Equation (1), $mg\cos{\theta}=N$, this satisfies observation $(2)$.

Along the surface: Equation (2),$ma=mg\sin{\theta}$, this satisfies observation $(2)$ and $(3)$.

These equations in combined sense, tell us that the net force on the block is $mg\sin{\theta}$. The normal force has cancelled out $mg\cos{\theta}$.

Therefore, this is the correct FBD.

Moreover, we also know this by observation that the block will move both forward and downward. So, if you further resolve $mg\sin{\theta}$ vertically and horizontally, you will get the true vertical and horizontal acceleration of the block respectively.

However, if the wedge is accelerating too, you just have to form equations that satisfy the observations. The most important observation in either of the cases is the block is stationary with respect to the wedge's surface.

I hope forming correct equations wouldn't be a problem for you anymore.

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    $\begingroup$ The first FBD is also correct. Setting up a FBD and solving the equations of motion is a universal method that works in any direction. Later in your answer you are correctly saying that there will be motion in horizontal and vertical direction, but your equations of motion for the first FBD are saying something different. Set up the equations correctly, and the problem can be solved also. You will need an additional equation though. You can state that the motion perpendicular to the slope is zero, like you did for FBD nr. 2. $\endgroup$
    – Orbit
    Commented Jan 24, 2018 at 10:18
  • $\begingroup$ I'm sorry, but it is still not correct. Please see the edit in my answer. $\endgroup$
    – Orbit
    Commented Jan 26, 2018 at 14:03
  • $\begingroup$ The inertia term in vertical direction should be $ma\sin\theta$, and in horizontal direction $ma\cos\theta$. $\endgroup$
    – Orbit
    Commented Jan 26, 2018 at 16:41
  • $\begingroup$ My equations are correct, perhaps we are talking about different things. $\endgroup$
    – Mitchell
    Commented Jan 26, 2018 at 20:01
  • $\begingroup$ For vertical equilibrium you are stating that there will be no acceleration in vertical direction, so it will not move vertically, and for horizontal equilibrium you are saying the block will accelerate away from the slope with acceleration a. You are saying the block will start to fly away from the slope, that does not make any sense. $\endgroup$
    – Orbit
    Commented Jan 27, 2018 at 13:51

Some context is needed. The answer depends on the acceleration.

  • Frictionless ramp: in this case, the block accelerates down the ramp. There is no acceleration in a direction normal to the ramp, so there is no net force in that direction, so $N=mg \mathrm{cos}\theta$.

  • Banked turn: in this case, the acceleration is horizontal, so there is no net vertical force. Thus $N\mathrm{cos}\theta=mg$.

  • $\begingroup$ That was exactly what was bugging me..... yeah I do wanna know how do I come to Different results for the same trigonometry.... Sorry for being a rough question asked but I thought providing specifics is just gonna make the question look as insincere $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 15:26
  • $\begingroup$ One has to draw FBD based on the motion that is observed. Not the other way around $\endgroup$ Commented Feb 9, 2018 at 6:34

The physical principle behind problems like this is Newtons first law:
Where F is the sum of all forces in a direction.

In most school problems, there is no motion( and thus acceleration) in the considered direction, and the equation reduces to $F=0$. Unfortunately many people jump to the conclusion that $F=0$ without motivating it. Many times they are lucky and there is indeed no motion in that direction, and they get the right answer. But sometimes the assumption that there is no acceleration is wrong. In that case they get the wrong answer, and may not be able to figure out why. This is a perfect example of a situation like that. Without friction, which is the case according to the FBD drawn, the block will accelerate down the slope. A component of this acceleration is in the vertical direction, thus the assumption that the sum of forces in the vertical direction is zero is incorrect. This results in a wrong answer.

It has been suggested in other answers that the FBD is incorrect and should be solved in certain directions. This is not true. The problem can be solved by decomposing the forces in any two directions, as long as they are not on the same line. The proof below shows that the problem can be solved by decomposing the forces along and perpendicular to the slope, but also in horizontal and vertical direction.

Let's start with the easy solution, and derive the equations of motion in the direction of the slope and perpendicular to it. I'll define the t direction as going down the slope, and the n direction perpendicular to it downward. The equations are:

$$\begin{align} mg\sin\theta &= ma_{t} \tag{1a} \\ mg\cos\theta - N &= ma_{n} \tag{2a} \end{align}$$

Now we should note that N is not a random force, but it is the reaction force that keeps the block on the slope. Therefore the acceleration perpendicular to the slope is zero, and we will call the acceleration along the slope a. The equations now reduce to:

$$\begin{align} mg\sin\theta &= ma \tag{3a} \\ mg\cos\theta - N & = 0 \tag{4a} \end{align}$$


$$\begin{align} ma &= mg\sin\theta \tag{5a} \\ N &= mg\cos\theta \tag{6a} \end{align}$$

Now the difficult part, solving the system in horizontal and vertical direction. We will define the x direction as to the left, and the y direction downward. We then get the following equations:

$$\begin{align} mg - N\cos\theta &= ma_{y} \tag{1b} \\ N\sin\theta &= ma_{x} \tag{2b} \end{align}$$

Again we know that the block can only move along the slope, because N is a reaction force, and again we call the acceleration along the slope (downward) a. So we get the following equations:

$$\begin{align} mg - N\cos\theta &= ma\sin\theta \tag{3b} \\ N\sin\theta &= ma\cos\theta \tag{4b} \end{align}$$

Equation 4b can now be rewritten as:

$$\begin{align} N &= \frac{ma\cos\theta}{sin\theta} \tag{5b} \end{align}$$

Now equation 5b can be inserted in 3b:

$$\begin{align} mg - ma \frac{cos\theta}{sin\theta} cos\theta &= ma\sin\theta \tag{6b} \end{align}$$

This can be rearranged to:

$$\begin{align} mg &= ma \big( \frac{cos\theta}{sin\theta} cos\theta + sin\theta \big) \tag{7b} \end{align}$$

Now we multiply the second term in the brackets with $\frac{sin\theta}{sin\theta}$ and get:

$$\begin{align} mg &= ma \big( \frac{1}{sin\theta} \big) \tag{8b} \end{align}$$


$$\begin{align} ma &= mg\sin\theta \tag{9b} \end{align}$$

This is the same as equation 5a. Inserting this in equation 5b gives:

$$\begin{align} N &= mg\sin\theta \frac{cos\theta}{sin\theta} \tag{10b} \end{align}$$

This simplifies to:

$$\begin{align} N &= mg\cos\theta \tag{11b} \end{align}$$

Which is again the same as 6a.

The important lesson to be learned here is to always write down the accelerations explicitly, and only to set them to zero after you prove they will be zero.

  • $\begingroup$ That’s not the point... $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 15:04
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    $\begingroup$ Your question was why your results are wrong, so I think it is the point. $\endgroup$
    – Orbit
    Commented Jan 23, 2018 at 15:09
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    $\begingroup$ Yeah I do get some of it how that makes a difference... Sorry for sounding harsh but a newbie makes the most noise you know... $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 15:28
  • $\begingroup$ Don't worry about it, i probably deserve it, done it prenty of times myself ;). Try to solve your equations paralell and perpendicular to the slope, then sum of forces is zero perpendicular, and you can get the acceleration or needed friction in the other direction. Also keep in mind that N should be smaller than mg. $\endgroup$
    – Orbit
    Commented Jan 23, 2018 at 15:42

you should not resolve the normal into components This is because Ncosθ≠mg. There is non-zero acceleration in the downward direction.if the block is in equilibrium there is some other force pushing the block along the inclined plane. the force acting along the inclined plane which is pushing the body up can be resolved into components

1.the normal force is due to contact between the body and the plane

2.the mg(downward force) is due to gravity

the force which is not normal to the surface is resolved

  • $\begingroup$ Let’s take a banked road at a fixed angle then,there it’s a premise to write cosine times N is equal to mg... $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 15:08

You have made a mistake in applying newton's 2nd law here. It says that $\sum \vec{F}=m\vec{a}$.

Since the body accelerates along the slope of the ramp (say, with acceleration $a$) there will be a component of acceleration $a\sin{\theta}$ which is along the vertical. So newton's 2nd law, when applied in the vertical direction becomes: $$mg-N\cos{\theta} = ma\sin{\theta}$$

Suppose instead, we apply newton's 2nd law along the direction of the normal reaction, then the component of acceleration along the normal is $a\cos{90^{\circ}} = 0$. So newton's 2nd law becomes: $$N - mg\cos{\theta}=ma\cos{90^{\circ}}=0$$ $$\Longrightarrow N=mg\cos{\theta}$$

Now, you can use both the FBD's to solve the problem, theres nothing wrong with that. Each free body diagram gives you 2 equations (along the 2 perpendicular directions), which you need to solve the 2 variables in the problem, $N$ and $a$. Normally, people use the first free body diagram, because then you don't have to solve the equations instead you directly can solve for the variables (as you have done)


Before summing forces as described in the other answers, I find it helpful to add a little more detail to the FBD. Here, it's easy to see that N = mg cos(theta) instead of mg = N cos(theta)


Here's a FBD for a banked road, with centripetal force shown.

enter image description here

  • $\begingroup$ Very explanatory and concise answer... $\endgroup$
    – Dilin Finn
    Commented Feb 8, 2018 at 13:49
  • $\begingroup$ One more favour,a banked road fbd,please:)? $\endgroup$
    – Dilin Finn
    Commented Feb 8, 2018 at 13:59
  • $\begingroup$ @DilinFinn, I added a banked road fbd $\endgroup$
    – BCott
    Commented Feb 10, 2018 at 6:25
  • $\begingroup$ This pair of FBD's looks equivalent, but they are not. For a block going down an inclined plane, N<mg. For the car on the banked curve, if the car is going just fast enough for the x component of the normal force to provide all of the centripetal force required for circular motion, N>mg. $\endgroup$ Commented Jan 18, 2019 at 3:48
  • $\begingroup$ @DavidWhite, I agree, they're not equivalent. The small green arrow parallel to Nmg in the second diagram depicts the additional normal force for a banked road.. Should I have depicted it differently? $\endgroup$
    – BCott
    Commented May 30, 2019 at 5:58

You made a very simple trigonometric mistake.

When you equated $N=mg \cdot cos(\theta)$ its wrong. Look up $cos(\theta)$ definition and you will understand.


noun, Mathematics

the trigonometric function that is equal to the ratio of the side adjacent to an acute angle (in a right-angled triangle) to the hypotenuse.

Based on this, the relation has to be, in the right top diagram, as $N=\frac{mg}{cos(\theta)}$

And indeed, you arrive at the same relation in your second diagram.

REMEMBER : Number never lie; if they appear to, then it must be our mistake :P

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    $\begingroup$ That's not what the OP is asking. $\endgroup$
    – Mitchell
    Commented Jan 29, 2018 at 10:57
  • $\begingroup$ @Mitchell. I identified my mistake. I'll be correcting it. But I stand by base opinion i.e. the issue is trigonometric in nature. OP has resolved the components wrongly. I had faced the same issue in my school days. That's why I could say this confidently! $\endgroup$ Commented Jan 29, 2018 at 12:02
  • $\begingroup$ Well,if you have two given lines,you can always cos theta the other with the first. That’s what I tell myself,and that works too. $\endgroup$
    – Dilin Finn
    Commented Feb 8, 2018 at 13:47
  • $\begingroup$ So,I get my mistake then(make a triangle and all) then can u show the fbd I have to make in the case of a banked road? $\endgroup$
    – Dilin Finn
    Commented Feb 8, 2018 at 13:48
  • $\begingroup$ @DilinFinn-I understand your doubt. My answer is wrong. I checked it. Now, for banked road FBD, Ncos(θ)=mg. But not for the case when a block slides along the wedge. The motion is different in both the cases, hence different free body diagram. The lateral component, that is Nsin(θ) will offer the centripetal force. $\endgroup$ Commented Feb 9, 2018 at 6:30

You are splitting the forces correctly. But why are you setting them equal?

You do that because you assume Newton's 1st law. The problem is just that Newton's 1st law doesn't apply in both cases. It only applies in the direction perpendicular to the surface, since nothing accelerates in that direction. So, your second equation is correct.

In any other direction, there is an acceleration component included, so you must use Newton's 2nd law and include a mass-times-acceleration-component term. The first equation is wrong and should have been:



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