# Newton's third law -- I have trouble with some real-life examples

• Juni M
In summary, forces always occur in pairs and are equal in magnitude but opposite in direction. According to the second law, force is the product of mass and acceleration. In real-life examples, when a stone is towed at a stone wall, the force exerted by the stone on the wall is equal to the stone's mass times its acceleration, which is the decreasing velocity as it comes to a standstill. The stone wall does change velocity as a consequence of the impact, but it is also affected by other forces, making its net acceleration small. In another example, if a person pushes on a wall or stands on the ground and comes to a standstill, the acceleration can be understood by considering the closed system of the Earth and the

#### Juni M

Homework Statement
Hi
I’m having a hard time getting my head around Newtons third law. I’m sorry if this question is long but I thought that by explaining how Iw understand it, and what I’m not understanding, it might be easier for someone to spot exactly what it is I’m not getting what piece of information I’m missing. Also, sorry if my spelling is faulty I am not a native english spacer.
Relevant Equations
F=ma
m1a1=m2a2 right?
I have gathered that forces always occur in pairs and are equal in magnitude but opposite in direction and that according to the second law a force is the product of the mas and acceleration of the object exerting the force.

My problem is with getting this make sense with real life examples. If I tow a stone at a stone wall, the force exerted by the stone on the wall is equal to the stones mas times its acceleration. (And by acceleration in this case I take to be the decreasing velocity as the stone comes to a standstill when it hits the wall. Right? ) But does the stone wall change velocity as a consequence of the stone hitting it ? Because if it is not moved by the impact, how can it have an acceleration, and without the acceleration how can it exert a force?

Or maybe it does have an acceleration but in that case it surely must be infinitesimally small so that multiplied even with the walls big mas it should yield a really small force. I am having a hard time believing that the acceleration of the wall can be big enough to make the force exerted from the wall be the same as that exerted from the stone, so I’m clearly missing something!

Another example is if I push on the wall org stand on the ground, as I come to a standstill pushing the wall or standing on the ground, how am I to understand the acceleration if both the wall, the ground and I apparently are not moving?
How am I to mace sense of this?

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PeroK
Juni M said:
But does the stone wall change velocity as a consequence of the stone hitting it ?
Yes. However, typically the wall is firmly anchored in the ground and also being affected by forces from there, meaning its net acceleration is not given only by the impact but also from other forces. Looking a bit bigger, the wall is anchored to the Earth, whose mass is enormous compared to the stone's mass and therefore will not have any noticeable acceleration (i.e., acceleration will be much smaller than what you could hope to measure). Also, you probably affected the Earth in the opposite direction when throwing the stone in the first place.

Lnewqban
Juni M said:
Homework Statement:: Hi
I’m having a hard time getting my head around Newtons third law. I’m sorry if this question is long but I thought that by explaining how Iw understand it, and what I’m not understanding, it might be easier for someone to spot exactly what it is I’m not getting what piece of information I’m missing. Also, sorry if my spelling is faulty I am not a native english spacer.
Relevant Equations:: F=ma
m1a1=m2a2 right?

I have gathered that forces always occur in pairs and are equal in magnitude but opposite in direction and that according to the second law a force is the product of the mas and acceleration of the object exerting the force.
My problem is with getting this make sense with real life examples. If I tow a stone at a stone wall, the force exerted by the stone on the wall is equal to the stones mas times its acceleration. (And by acceleration in this case I take to be the decreasing velocity as the stone comes to a standstill when it hits the wall. Right? ) But does the stone wall change velocity as a consequence of the stone hitting it ? Because if it is not moved by the impact, how can it have an acceleration, and without the acceleration how can it exert a force? Or maybe it does have an acceleration but in that case it surely must be infinitesimally small so that multiplied even with the walls big mas it should yield a really small force. I am having a hard time believing that the acceleration of the wall can be big enough to make the force exerted from the wall be the same as that exerted from the stone, so I’m clearly missing something!
Another example is if I push on the wall org stand on the ground, as I come to a standstill pushing the wall or standing on the ground, how am I to understand the acceleration if both the wall, the ground and I apparently are not moving?
How am I to mace sense of this?
To make sense of the laws of mechanics you really need a closed system. Imagine the scenario where the Earth is perhaps not so large as it is and you are the only person on it. And it's not spinning. You have a ball in your hand and throw it at a wall:

1) To throw the ball you apply a force to the ball, and by Newton's third law the ball exerts an equal and opposite force to you, which is transmitted by your feet to the ground.

2) As the ball is flying towards the wall, the Earth will be moving and rotating slowly in the opposite direction.

3) When the ball bounces off the wall and reverses its direction the the Earth will reverse its motion as well. Again by Newton's third law. And, as the ball is flying back to you, the Earth will be moving and spinning slightly in the direction of the wall.

4) When you catch the ball the force to stop the ball will be exactly equal to the force to stop the Earth moving and spinning.

5) Everything will be back at rest as it was before you threw the ball.

As mentioned above, the Earth is far too big for us to notice this under normal circumstances.

Juni M and Lnewqban
Orodruin said:
Yes. However, typically the wall is firmly anchored in the ground and also being affected by forces from there, meaning its net acceleration is not given only by the impact but also from other forces. Looking a bit bigger, the wall is anchored to the Earth, whose mass is enormous compared to the stone's mass and therefore will not have any noticeable acceleration (i.e., acceleration will be much smaller than what you could hope to measure). Also, you probably affected the Earth in the opposite direction when throwing the stone in the first place.
Thank you ! I think I understand that part now. So I can think of it like indirectly throwing the stone at earth, and sins the mas of Earth is so big even though the acceleration it would get from the impact with the stone is unmeasurably small, the product would yield a force equal to that exerted by the stone!?

Sometimes doing a quick calculation and putting in some numbers can be informative. Let's say you release a 1-kg stone from a certain height so that it hits the ground in 1 sec. Assuming that the acceleration of gravity is rounded to 10 m/s2, you can calculate the height from which the stone was dropped as follows:$$h=\frac{1}{2}gt^2=\frac{1}{2}\times 10 ~\text({m/s}^2)\times 1^2(\text{s}^2)=5~\text{m}.$$ You are a believer in Newton's third law so you use the equation ##m_1a_1=m_2a_2## to find the acceleration of the Earth and from it the distance it travels in the same 1 sec:
$$m g=M_{\text{Earth}}a_{\text{Earth}}\implies a_{\text{Earth}}=\frac{m}{M_{\text{Earth}}}g=\frac{1~(\text{kg})}{6\times 10^{24}~(\text{kg})}\times 10 ~(\text{m/s}^2)=1.7\times 10^{-24}~(\text{m/s}^2).$$Now for the distance, $$h_{\text{Earth}}=\frac{1}{2}\times 1.7\times 10^{-24}~(\text{m/s}^2)\times 1^2(\text{s}^2)=0.85\times 10^{-24}~(\text{m}).$$ To compare with something small, that distance is about 100 trillion times (1014) smaller than the diameter of an atom.

The above is essentially @Orodruin's argument in post #2. I put in the numbers to make it stand out. The take-home message is that the distance by which the Earth moves is undetectable so one is led to believe that only the dropped object moves.

PeroK said:
To make sense of the laws of mechanics you really need a closed system. Imagine the scenario where the Earth is perhaps not so large as it is and you are the only person on it. And it's not spinning. You have a ball in your hand and throw it at a wall:

1) To throw the ball you apply a force to the ball, and by Newton's third law the ball exerts an equal and opposite force to you, which is transmitted by your feet to the ground.

2) As the ball is flying towards the wall, the Earth will be moving and rotating slowly in the opposite direction.

3) When the ball bounces off the wall and reverses its direction the the Earth will reverse its motion as well. Again by Newton's third law. And, as the ball is flying back to you, the Earth will be moving and spinning slightly in the direction of the wall.

4) When you catch the ball the force to stop the ball will be exactly equal to the force to stop the Earth moving and spinning.

5) Everything will be back at rest as it was before you threw the ball.

As mentioned above, the Earth is far too big for us to notice this under normal circumstances.
Thank you, this was very clarifying !

berkeman
kuruman said:
Sometimes doing a quick calculation and putting in some numbers can be informative. Let's say you release a 1-kg stone from a certain height so that it hits the ground in 1 sec. Assuming that the acceleration of gravity is rounded to 10 m/s2, you can calculate the height from which the stone was dropped as follows:$$h=\frac{1}{2}gt^2=\frac{1}{2}\times 10 ~\text({m/s}^2)\times 1^2(\text{s}^2)=5~\text{m}.$$ You are a believer in Newton's third law so you use the equation ##m_1a_1=m_2a_2## to find the acceleration of the Earth and from it the distance it travels in the same 1 sec:
$$m g=M_{\text{Earth}}a_{\text{Earth}}\implies a_{\text{Earth}}=\frac{m}{M_{\text{Earth}}}g=\frac{1~(\text{kg})}{6\times 10^{24}~(\text{kg})}\times 10 ~(\text{m/s}^2)=1.7\times 10^{-24}~(\text{m/s}^2).$$Now for the distance, $$h_{\text{Earth}}=\frac{1}{2}\times 1.7\times 10^{-24}~(\text{m/s}^2)\times 1^2(\text{s}^2)=0.85\times 10^{-24}~(\text{m}).$$ To compare with something small, that distance is about 100 trillion times (1014) smaller than the diameter of an atom.

The above is essentially @Orodruin's argument in post #2. I put in the numbers to make it stand out. The take-home message is that the distance by which the Earth moves is undetectable so one is led to believe that only the dropped object moves.
Thank you so much for taking the time to write all of this out, this helps!

berkeman and kuruman