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I normally find that I can explain equal and opposite forces not having the same result in Newton's third law because those forces are acting on objects with different masses. However, this doesn't work for the following situation, and I haven't managed to resolve the seeming paradox in it yet.

Say you had a horse of 1 kg, pulling a cart of 1 kg. The horse exerts 5 N forward on the cart, and the cart exerts 5 N backward on the horse (I would draw a diagram, but I don't know how to with this software: sorry!). The horse is succeeding in pulling the cart forwards. If the horse is accelerating forwards with the cart, then there must be a resultant forward force on the horse. If you said that the horse was pushing on the ground where the horizontal component of his force was 10 N backwards, meaning the horizontal component of the ground's force on the horse is also 10 N forwards, then you could argue that, with 5 N backwards from the cart and 10 N forward from the ground, there is a resultant force of 5 N forwards on the horse. However, the horse can only generate the pulling force of 5 N on the cart by using this forward force that the ground is exerting on him/her, so surely this means that both of these forward forces must be equal? In which case, doesn't this mean that the ground could only be exerting 5 N forward on the horse if the horse was only managing to exert 5 N forward on the cart, meaning that when you factor in the 5 N backwards that the cart is exerting on the horse, the resultant force on the horse is 0, and it is impossible for the horse to accelerate the cart? And since it isn't impossible, what have I not factored in/reasoned wrongly here?

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  • $\begingroup$ Please note that the cart is also applying force on the ground and the ground too is applying a reaction force on the cart, a component of which gives too gives a forward push to the cart and the horse system. $\endgroup$
    – Agrim Arsh
    Aug 16, 2021 at 9:33
  • $\begingroup$ However, the horse can only generate the pulling force of 5N on the cart by using this forward force that the ground is exerting on him/her, why? $\endgroup$
    – ACB
    Aug 16, 2021 at 9:41
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    $\begingroup$ A horse of 1 kg. ... a Bonsai horse? :-) $\endgroup$
    – GiorgioP
    Aug 16, 2021 at 9:50
  • $\begingroup$ "However, the horse can only generate the pulling force of 5 N on the cart by using this forward force that the ground is exerting on him/her, so surely this means that both of these forward forces must be equal?" Is disagree with the "so surely" here. If only 5 N is needed to pull the cart with the given acceleration, then that is all that needs to be exerted. The force exerted on and from the ground is unrelated to this. $\endgroup$
    – Steeven
    Aug 16, 2021 at 11:39

2 Answers 2

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However, the horse can only generate the pulling force of 5N on the cart by using this forward force that the ground is exerting on him/her, so surely this means that both of these forward forces must be equal?

Your mistake is here in this part. The force that the horse exerts on the ground is not generally equal to the force that the horse exerts on the cart. It can be equal, higher, or lower. Assuming ground friction acts on the cart then the traction force on the horse could be equal, for example, if they are going at a steady speed, less if they are decelerating, and more if they are accelerating.

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There is a problem with this part

doesn't this mean that the ground could only be exerting 5N forward on the horse if the horse was only managing to exert 5N forward on the cart

The ground exerts 10N of forward force on the horse, but this 10N has to accelerate the two objects, the horse and cart.

Using the numbers in the question and $F=ma$ for the combined 'horse and cart'

$$10=2a$$

so $a = 5ms^{-2}$

On the horse alone, the resultant force is 10-5 = 5N, from $F=ma$ again

$$5=1a$$ $a=5ms^{-2}$, and for the cart alone

$$5=1a$$ $a=5ms^{-2}$

So everything can accelerate together at the same rate.

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