# Looking for an intuitive understanding of normal force

I understand normal force to be the perpendicular force to a surface of contact. However, I have come across a problem which has caused me to rethink this.

My initial understanding of force is demonstrated by the following diagram of an object sliding down an inclined plane. Here you can see that $F_N = F_g * cos(\theta)$

Now look at this banked curve problem. This is taken directly from Wikipedia here. It says that $F_N = F_g/cos(\theta)$ which is bigger than in the first scenario.

Why is the normal force different here?

The normal force comes from Newton's third law, as a reactive force. Before there is any centripetal accel, $F_{N}$ was $F_{g}∗cosθ$, but after there is centripetal accel, $F_{N}$ changes to $F_{g}/cosθ$. This makes $F_{N}$ larger in the second scenario even though there is no more force acting in the direction of the ground. What causes the increase in $F_{N}$? Something must push against the ground in order for $F_{N}$ to push back.

My thoughts and attempts at solving this:

There has to be a force acting in the direction of the road in order to increase the value of the normal force. I think this force comes from the fact that the car's velocity is causing a force on the road, pushing in to the road because the car's velocity wants to go straight, but the road is shaped in a curved bank so the car cannot go straight; the normal force of the road pushes back, which allows for the centripetal acceleration.

I've tried accounting for the difference in $F_{N}$ and I've found that $F_{N}$ in the second scenario is $Fg*cos\theta + Fg*tan\theta*sin\theta$ the difference being the second term. I haven't been able to describe the difference in $F_{N}$ in any other way. I believe the difference in $F_{N}$ should should somehow be related to the centripetal acceleration.

I think the normal force is a result of the car's (the ball in this case) velocity against the circular road causing the road to push back. I lack the mathematical ability to describe it.

• In the particular situation you're considering, note that the ball is accelerating. Whereas in the first case you have a static situation --- one in which the net force on the ball must be zero --- in the second the ball is moving in a circle. This continual change of direction requires a certain force to maintain. This force can come from two places --- gravity and the normal force. These must combine in such a way not as to make zero, but as to make some inwards pointing force that causes the ball to accelerate. – gj255 Jan 10 '14 at 21:42
• Actually I think I've completely missed the source of your confusion. What I've said above is true, but perhaps not relevant. In the first case, the equation you get arises from looking at the components of the forces along a line perpendicular to the plane. You know that there is zero motion in this direction (the block isn't levitating or sinking into the plane) so you can set this force to zero. In the second case, you're looking at the components of the forces in the vertical direction. We can set this to zero since the height of the ball above the ground is constant by assertion. – gj255 Jan 10 '14 at 21:49
• What I don't understand is how they derive $F_{N}*cos(\theta)=F_{g}$ I always thought the normal was the perpendicular force on a surface, but here it seems it is not calculated as such. I think it is somehow due to the acceleration inwards causing additional force into the banked road, which in turn increases the normal force. – Klik Jan 10 '14 at 22:45
• I want to understand how to know when the normal force changes like this. It is not equal to what I thought it would be equal to. Not just for centripetal force equations, but all equations. – Klik Jan 10 '14 at 22:46
• OK, let me try to write you an answer. – gj255 Jan 10 '14 at 23:22

You have asked what causes the increase in force from the earlier case, try seeing it from the body's frame as it will be in rest there. The body exerts a centrifugal force on the plane along with gravitational force, the resultant of these two forces is matched up by the normal provided by the banked road, now since the resultant of centrifugal force and gravitational force is more than gravitational force alone, the normal force increases in magnitude.

In terms of equation :

$F_g \cos(\theta) + m a_c \sin(\theta) = F_n$
$F_g \sin(\theta) = m a_c \cos(\theta)$

But if you see from ground frame the normal partly provides the centripetal force and also cancel outs the gravitational force i.e.

$F_n \cos(\theta) = F_g$
$F_n \sin(\theta) = m a_c$

In both frames as normal is involved in dealing with both forces it is greater than just the component of gravitational force.

• I am quite frustrated with myself. If you notice the comment I made on Carl's answer, I did mention that I thought $F_{N}=F_{g}*cos\theta + m*a_{c}*sin\theta$. I dismissed that idea however when I tried using geometry software to accurately graph the vectors and the answer did not match up. Seeing your answer made me take a second look at it and it helped a lot actually. I can't believe I overlooked the answer like that. Well, in any case, thanks for this answer. – Klik Jan 13 '14 at 2:50

If you're sliding down the plane, the normal force is the fraction of gravity which acts on the you. If you're moving in a banked curve, the normal force is that fraction of your centripetal force which is keeping you from sliding down the bank. I suspect that's the source of your discrepancy.

• I uploaded the images from my computer, I will use links from the web instead. Thanks for having a look! I am trying to draw this out accurately on paper as we speak to account for it. I think I've figured it out, but I want to prove it first before I am sure. I think the normal force is the sum of the $F_g *cos(\theta)$ and $ma*sin(\theta)$. – Klik Jan 10 '14 at 20:15
• @Klik I appreciate you trying to accommodate Carl's inability to view the images, but it's slightly preferable for images to be hosted on Stack Exchange's dedicated imgur subdomain where possible. That way we don't have to worry about them disappearing if external sites go down in the future. I've rolled back the question accordingly. Carl, perhaps you can come back and edit this later when you are able to see the images. – David Z Jan 10 '14 at 20:40
• Just wanted to mention that I did not end up figuring it out. So feel free to have a shot at it if you have time. – Klik Jan 11 '14 at 3:39

Suppose I placed a block on a table and lent on the top of it. There would be a downwards force on the block due to its weight, and a further downwards force on the block due to my arm. But the block wouldn't be moving --- it wouldn't be accelerating in the slightest. And so, by Newton's second law, there must be no force acting on it! What this means is that the table must be exerting a force equal in magnitude to the sum of both the weight of the block and the force from my arm, but upwards, so as to cancel everything out.

Hopefully this is familiar to you. The point of all this is that normal forces (and more generally, constraint forces) vary in just such a way as to maintain a certain condition --- in this case, it's the condition that the block doesn't sink into the table. You might be used to $N = mg$ or something similar, but that doesn't have to be so. Indeed, if I attached a string to the top of the block and pulled slightly on the end, the normal force would be less than $mg$. This is because we already have an upwards force acting on the block --- that due to the string --- and so the table doesn't have 'as much ground to make up' when it comes to making sure all the forces are cancelled.

With this said, let's approach the two systems you're looking at. You asked: "why is the normal force not $F_g \cos \theta$ like it was in the last problem?". I could ask you: why should it be the same? The two systems are rather different: one involves a stationary block, the other a ball undergoing circular motion, and we've seen that normal forces can vary from circumstance to circumstance. A priori there's no reason to assume that the normal forces should have the same form. But let's consider specifically why they don't.

In the first case, the block isn't moving and so (since its acceleration is therefore zero) the net force is zero. Or at the very least (suppose that in actual fact the block was sliding along the plane) we can say that the block isn't moving in the direction perpendicular to the plane, and so the forces in that direction add to zero. The component of the force of gravity in this direction is $F_g \cos \theta$, whereas the component of the normal force in this direction (which is just the total normal force) is $F_N$. These forces must cancel, and so

$$F_N - F_g \cos \theta =0\,.$$

In the second case, the block is moving and in fact it is accelerating. It's accelerating because it's constantly changing direction. It might seem counter-intuitive, but the direction of this acceleration is actually towards the centre of the circle. What this means is that the components of the forces in the horizontal direction won't cancel; they would only cancel if our acceleration were zero in this direction. As I've said, and as the diagram you gave suggests: there is acceleration in this direction.

However, there isn't any motion in the vertical direction. This is true essentially by assertion --- we're considering a problem in which a ball rolls around a banked curve at constant height. If there's no motion in the vertical direction, the components of the forces in the vertical direction must sum to zero. We see from the diagram that the only forces acting are the normal force and the weight of the ball. So, resolving the vectors correctly, we find that we must have

$$F_N \cos \theta - F_g = 0$$

if indeed the vertical forces are to sum to zero, as we require. The other equation --- that for components in the horizontal direction, will look like this

$$F_N \sin \theta = ma \,.$$

So to check we understand: if we were to resolve forces perpendicular to plane in this second example, we couldn't conclude that the forces we obtained summed to zero. The reason? Well, the ball has an acceleration that points in the horizontal direction, and so the component of its acceleration along the direction perpendicular to the plane is non-zero. As such, the net force in this direction is non-zero. It may not look like the ball has a component of acceleration perpendicular to the plane --- that would seem to suggest that the ball was trying to lift off! But, despite how it might seem, the fact of the matter is that the ball has an acceleration that points horizontally, and so there is a component of this acceleration along the direction perpendicular to the plane.

Hope this helps!

• You've described normal force as a force that pushes back. What I find difficult to understand, in the second scenario, the normal force doesn't depend on the force of gravity. It only depends on the inward acceleration. But, isn't gravity pulling the car downwards? Please correct me if I'm wrong. I was really hoping to find a way to calculate $F_N$ as a function of other forces, like gravity. – Klik Jan 11 '14 at 0:10
• I thought I had figured it out again, but to no avail. I appreciate your answer, but I've decided to leave this question unanswered until a proof to explain the difference in the Normal force is given. I believe it involves a calculus approach. All explanations I've found so far say N changes so the forces are balanced. However, I am looking for a deeper reasoning than for the sake of balancing the forces. I've seen many other people ask this question on the web and none have provided a good answer, so hopefully this may be one place where an answer can be found. – Klik Jan 11 '14 at 3:33
• So firstly, there is certainly no calculus involved in this problem --- it's vector algebra. The normal force in the second scenario does depend on the force of gravity, as you can see from the second equation --- $F_N \cos \theta = F_g$. Rearranging, $F_N = F_g /\cos \theta$. So you see, it does depend on gravity! I'm not sure what kind of explanation you're looking for, but balance of forces (or imbalance, as the case may be) is precisely what's going on here. We look at the motion of the body, and through $F = ma$, this tells us what the forces must be. Zero motion means zero force. – gj255 Jan 12 '14 at 11:43
• I realize there is no calculus involved in the solution of this problem. I understand how to solve this problem, but what I would like to know is how $F_{N}$ is derived. The normal force comes from Newton's third law, as a reactive force. Before there is any centripetal accel, $F_{N}$ was $F_{g}*cos\theta$, but after there is centripetal accel, $F_{N}$ changes to $F_{g}/cos\theta$. This makes $F_{N}$ larger even though there is no more force acting in the direction of the ground. What causes the increase in $F_{N}$? Something must push against the ground in order for $F_{N}$ to push back. – Klik Jan 12 '14 at 19:58
• Ah OK, I think I see. So indeed, if the ground is exerting a force $F_N$ on the ball, then the ball must be exerting a force $F_N$ on the ground. This force is greater in the second case than the first --- that is, the ball is exerting more force on the ground in the second case. But the force the ball exerts on the ground due to its weight is the same in both cases. So where does the extra force come from? Is this essentially what you're asking? – gj255 Jan 12 '14 at 20:24