# Newton's third law and a horse

1. Jan 3, 2015

### oreo

When a horse pulls tonga, the Tonga exerts an equal force on Tonga then why does it move. How is Newton's third law applied to such case.

2. Jan 3, 2015

### PWiz

The tonga exerts a force on the horse, not on itself. Action and reaction forces, though equal and opposite, act on different objects and hence do not cancel each other.

3. Jan 3, 2015

Short answer: the tonga has a smaller mass, and so it would experience a larger acceleration.

Long answer: Newton's third law states that when two bodies interact, they exert equal and opposite forces on each other. Expressed mathematically in vector form:

$$\vec{F}_{12} = -\vec{F}_{21}$$

Where $\vec{F}_{AB}$ means the "force of A on B".

$$\vec{F}_{horse,tonga} = -\vec{F}_{tonga,horse}$$

Ignoring the forces of friction and air resistance we have by Newton's second law:

$$m_{tonga} \vec{a}_{tonga} = -m_{horse} \vec{a}_{horse}$$

In terms of magnitudes:

$$m_{tonga} {a_{tonga}} = m_{horse} {a_{horse}}$$

The less massive body (in this case the tonga) gets a larger acceleration from the interaction.

Last edited: Jan 3, 2015
4. Jan 3, 2015

### Satwik

You have not considered all the forces here.
The road exerts frictional force due to which horse moves forward along the cart and frictional force is also exerted on the cart .
Now both have the same acceleration because friction is a self adjustable force

5. Jan 3, 2015

### oreo

You are saying that the Tonga is experiencing the greater acceleration, so it moves. But according to your answer, there is net acceleration which in turn means that there is net force on Tonga which can't happen.

6. Jan 3, 2015

### oreo

How would you answer if both have same masses.

7. Jan 3, 2015

### Satvik Pandey

Hi Shayan. Draw the Free body diagrams first.
Assume the red block to be a cart and the black box to be horse (Just imagine). I have mentioned the force acting on the horse and cart. I have not drawn normal and frictional force acting on the horse and cart. You can see that a $net$ force acts on the cart. That's why it moves.

8. Jan 3, 2015

### Staff: Mentor

Yes, it can happen. Why wouldn't it be able to happen? Third law pairs act on different objects.

9. Jan 3, 2015

### FactChecker

Don't forget that the horse is also pushing against the Earth, so there is an imperceptible motion of the earth. That accounts for the combined horse and Tonga moving in the opposite direction.

10. Jan 3, 2015

### oreo

Thank you. You are right. This simple because the forces are acting on two different bodies so they do not cancel each other. But according to MohammedRadey97 the less massive body would experience net force which means if cart is more massive then it would pull horse backward. I think so that there is some different explanation.

11. Jan 3, 2015

### sophiecentaur

You seem to have more or less got this now. The fact is that N3 says nothing about Equilibrium and it's only when there are balanced forces (each with their own reaction force btw) that you get equilibrium and no acceleration.

12. Jan 3, 2015

### my2cts

The sum of all forces in a closed system vanishes.
Corollary: if the sum of forces does not vanish then the system is not closed. Find the missing parts and start over.

13. Jan 3, 2015

### PeroK

The key is the force of the horse pushing against the ground. If you were out in space and pulled something with a rope, then you would move towards the object you are pulling as well as it moving towards you. By Newton's third law, both you and the object would experience the same force; and, by the second law, the acceleration of each would be proportional to the mass. So, if you pulled a more massive object, you would accelerate faster than it.

On the Earth, the horse is pushing against the ground with a greater force. This force effectively accelerates both horse and cart.

On the TV in the UK recently there was the "World's Strongest Man" competition. One test involves pulling a giant truck. To do this, the strongman must exert a huge force using friction on the ground (they all wear rock-climbing shoes as these have the greatest friction). Most of this force is transmitted to the truck through a rope, and the truck pulls back on the man, so there is only a small nett acceleration of both.

In short, there are two forces acting on the horse and only one on the cart.

14. Jan 3, 2015

### oreo

Thank you. You have presented a logical explanation for this and cleared my confusion. I was also thinking of applying this law in space and imagined the same thing. Thanks again.

15. Jan 3, 2015

### oreo

T
Thanks for correcting my shortcoming.

16. Jan 3, 2015

I intentionally avoided other forces because the point I was trying to make was that Newton's third law does not imply equilibrium. Given two masses in vacuum, the larger mass always "wins", so to speak, inspite of the fact that there are equal and opposite forces between the masses.
Also, when I mentioned acceleration, I was explaining why, at least in space, it is tonga that moves towards the horse, and not the other way around. This has nothing to do with the acceleration of the system as a whole as internal forces in a system do not appear in Newton's second law (when applied to that system).

Last edited: Jan 3, 2015
17. Jan 3, 2015

### Andrew Mason

Does the horse not accelerate and, therefore, move toward the tonga? Think of the motion of the two bodies in relation to an inertial frame of reference (eg. their centre of mass).

AM

18. Jan 3, 2015

It does accelerate towards the tonga, bit the tonga's acceleration towards the horse is greater, right?

19. Jan 4, 2015

### Andrew Mason

If the tonga has less mass than the horse (I have no idea what a tonga is) it will accelerate toward the centre of mass (com) at a higher rate than the horse accelerates toward the com. They will collide before either one actually reaches the com.

If the separation is d, the com is located at a point (0) if the location of the horse is (-dmT/(mH+mT) and the location of the tonga is (dmH/(mH+mT). In time t each of the bodies will move a distance 1/2at2 toward the com (i.e. the distance a body moves is proportional to the body's acceleration). Since a=F/m and the forces are equal in magnitude, the distance that each moves is inversely proportional to its mass.

AM

Last edited: Jan 4, 2015
20. Jan 4, 2015