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$R$ = 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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Steeven
    Commented Jul 5, 2017 at 12:40
  • $\begingroup$ Sorry. I completely forgot to label the forces - I will do so now. $\endgroup$ Commented Jul 5, 2017 at 12:47
  • $\begingroup$ I've amended the question. Sorry for the confusion. $\endgroup$ Commented Jul 5, 2017 at 12:50
  • $\begingroup$ Is the object moving so that we are talking about kinetic friction? $\endgroup$
    – Steeven
    Commented Jul 5, 2017 at 12:51
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Jul 5, 2017 at 13:07

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jul 5, 2017 at 13:15
  • $\begingroup$ $R$ 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. $\endgroup$
    – Javier
    Commented Jul 5, 2017 at 13:59
  • 3
    $\begingroup$ 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. $\endgroup$
    – Javier
    Commented Jul 5, 2017 at 14:11
  • 2
    $\begingroup$ 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. $\endgroup$
    – Javier
    Commented Jul 5, 2017 at 14:12
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jul 5, 2017 at 15:23

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.


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