Why doesn't limiting friction contradict Newton's Third Law?

Question

$R$ = Normal reaction force exerted on the particle by the ground

$Fr$ = Frictional contact force between the particle and the ground

All forces are measured in Newtons ($N$)

Using the angles we can calculate the forces acting on the particle.

$$m=1, \ \ \ mg = g = 9.8N$$

Netwton's Third Law of motion states that if A exerts a force magnitude N on B, then B exerts a force of maginitude N on A, but in the opposite direction. Hence:

$$R = 9.8 + 30 \ sin(30)$$ $$R = 24.8N$$

Now, if we also go by Newton's Third Law, then if the particle is pushing forwards on the ground with a force magnitude N, then the ground is pushing backwards on the particle with a force magnitude N, which should be friction. However:

$$30 \ cos(30) = 26.0N$$ $$Fr_{(max)} = R \mu = 2.48N$$ $$Fr_{(max)} < 26.0$$

Due to the limit on friction set by the coefficient of friction between the surface and the particle, the particle accelerates with acceleration $a$. However, this appears to contradict Newton's Third Law, which clearly states that $Fr$ should equal $30 \ cos(30)$.

This is especially strange since $R$ follows Newton's Third Law perfectly without a limit, so why does friction appear to disobey it?

What am I missing here?

What is $Fr$? What is the $R$ in the $R\mu$ term? By the way, it is a bit unfortunate to use N as a dummy number to show examples, when also normal force is $N$ and the unit of Newton's is $\mathrm N$. I am getting a bit confused while reading. — Steeven, Jul 05 '17 at 12:40
Sorry. I completely forgot to label the forces - I will do so now. — Pancake_Senpai, Jul 05 '17 at 12:47
Is the object moving so that we are talking about kinetic friction? — Steeven, Jul 05 '17 at 12:51
Yes the object is in motion now that the 30N force is applied. It has acceleration = $a$ in the direction shown on the diagram. Prior to that ithe particle would have been at rest because the net force would have been 0N. — Pancake_Senpai, Jul 05 '17 at 12:57
All the forces in your diagram appear to be acting on one object, the sphere. How can you expect Newton's third law (which concerns forces on different bodies) to be relevant – let alone contradicted? — Philip Wood, Jul 05 '17 at 13:07

score 2 · Accepted Answer · answered Jul 05 '17 at 13:05

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One word on notation: in general, you should put units everywhere, and not just now and then. An equation like $30 \cos(30)=26.0N$ is unacceptable; either state that you're going to measure all forces in $N$ and don't bother writing the $N$, or write it everytime.

Okay, so onwards to the answer. First, your calculation of the normal force is correct but it's not because of Newton's third law, it's because of the second. You simply set the sum of vertical components equal to zero.

The problem in your reasoning is that the force that the object exerts on the ground is not the horizontal component of the pushing force. The only possible horizontal force between the object and the ground is friction, so the action-reaction pair is just the two friction forces the object and the ground exert on each other. According to your calculations, the frictional force turns out to be $2.48\, \mathrm{N}$ (actually we would need the dynamic coefficient but let's keep it simple). Therefore, your object exerts a force on the ground with horizontal component equal to $2.48\, \mathrm{N}$ in the forward direction.

How you can you tell that the $30\, \mathrm{N}$ force is unrelated to the ground? Well, look at your diagram. That force is being exerted by your hand (presumably) and it acts on the object, so that is the pair you care about. If you want to apply Newton's third law to the object-ground system, look at the picture again: the forces being exerted by the ground are the normal force and friction. Those are the ones you care about.

answered Jul 05 '17 at 13:05

Javier

28,106

I understand from your answer that the 30N force is not acting between the particle and the surface, hence the frictional force is not matching it (because the frictional force's force pair is the equal in magnitude but opposite in direction frictional force, also between the particle and the ground). However, why then does R balance out all downwards vertical forces. The normal reaction is between the particle and the ground, so why should the 30N force effect it? – Pancake_Senpai Jul 05 '17 at 13:15
$R$ balances the vertical forces because, since the object is not moving vertically, its vertical acceleration is zero. Therefore, the sum of all vertical components of forces is zero. – Javier Jul 05 '17 at 13:59
Doesn't that require the assumption that the object is not going to move vertically? If you were asked to calculate the direction of motion of the object, you would have to assume that there is no vertical motion to apply that logic. Also, don't the 2nd and 3rd laws contradict in this case? What is the reaction pair force with R? If it is the weight of the object only, then by increasing R is not balanced, which disobeys the 3rd law. If it is the net vertical force as transmitted through the object, then why is the reaction pair of Fr not the net horizontal force? – Pancake_Senpai Jul 05 '17 at 14:03
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Indeed you have to assume that. The ground has a maximum force it can withstand before collapsing, and if the pushing force is smaller than that, the object won't move vertically. R is a force exerted by the ground on the object, so its pair is a force exerted by the object on the ground, and it has the same magnitude. It doesn't appear in your diagram because you only drew the forces acting on the object. – Javier Jul 05 '17 at 14:11
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Again, the reaction force is equal to the total downwards force on the object because of the second law, not because of the third, and it's only because we're assuming the object doesn't move vertically. There is no reason to expect the horizontal forces will be balanced. – Javier Jul 05 '17 at 14:12
R is counterbalancing the net vertical force, yet Fr isn't counterbalancing the net horizontal force. Why does this occur according to the THIRD law? I understand the second law, but its the third law I'm struggling with. – Pancake_Senpai Jul 05 '17 at 14:39
The third law doesn't say that all forces must be balanced; it says that if object A exerts a force on object B, the object B exerts an equal an opposite force on object A. Having two objects is essential. In your example, the vertical force on the object is balanced because of the second law, not because of the third law. Since there's no assumption that the object doesn't move horizontally, there is no reason the horizontal forces should be balanced. See https://physics.stackexchange.com/questions/45653/with-newtons-third-law-why-are-things-capable-of-moving – Javier Jul 05 '17 at 14:54
The 3rd Law tells you how the forces on two different objects compare with each other due to the singular interaction of those two objects. The ball touches the ground, therefore, the ground exerts a force on the ball AND the ball exerts an equal and opposite force on the ground. There might be other forces from other interactions. Those other forces affect the acceleration, based on 2nd Law. The 3rd Law tells us nothing about the acceleration or about other forces that might act on an object. 3rd Law doesn't apply to the net force, only to the singular direct interaction. – Bill N Jul 05 '17 at 15:05
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It's clicked. I understand it now. The object's weight does NOT act on the surface in any way, and neither does the $30sin(30)$ force. They both act on the object to pull/push it down. This force is transmitted by the ball to the table, so the table pushes back on the object with force R. Meanwhile, the $30cos(30)$ force pushes horizontally on the object. This has NO impact on the ground. However, when the object moves right it pushes the surface to the right (friction), so the surface responds by pushing the object to the left. – Pancake_Senpai Jul 05 '17 at 15:23
Thank you both for all your help Javier and @Steeven. I finally "get" the third law. I would accept both your answers if I could but I guess I'll just flip a coin and see who gets the big green tick. – Pancake_Senpai Jul 05 '17 at 15:26

score 2 · Answer 2 · edited Jul 05 '17 at 20:30

First of all, you don't need to write $_{max}$ or talk about limits when it is kinetic friction. Kinetic friction has a fixed formula. That formula is:

$$F_r=\mu N$$

and not $F_r=\mu R$. Don't use the reaction force $R$, only the normal force $N$.

Now, this friction is what the ball affects the surface with. It is not the reaction force, which affects the surface, it is the friction which does.

The ball affects the surface with friction, and the surface holds back with the same force in the ball, yes. This is Newton's 3rd law.

this appears to contradict Newton's Third Law, which clearly states that $F_r$ should equal $30 \cos(30)$.

This is not correct, as I stated just above. The $30 \cos(30)$ is the reaction force's horizontal component $R_x$, but that is not the force that affects the surface. This $R_x$ causes friction, you could say, but they are not equal - Newton's 2nd law says that:

$$R_x-F_r=ma\quad\Leftrightarrow\quad F_r=R_x-ma$$

The $R_x$ is clearly larger than the friction $F_r$; the rest of $R_x$ Is used for acceleration.

Why doesn't limiting friction contradict Newton's Third Law?

2 Answers2