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When a horse pulls a cart, suppose connected by a string, it produces a tension $T$ in the string. The cart in turn pulls the horse by an equal force $T$ in the opposite direction. If this T>f, the friction on the cart, it moves forward. But how does the horse move with the cart pulling it? From what i have understood the horse presses the ground in an inclined direction and the horizontal component of the corresponding reaction is the force with which it pulls the cart i.e the tension T. So the more effort it puts, the same magnitude of force will pull it back. I do have the intuition for Newton's third law, that action and reaction are on different objects, for eg incase of a gun firing a bullet, the bullet exerts the same force to the gun, but the gun is much heavier so it can only nudge the gun a bit in the opposite direction. It makes sense. But I'm having a bit of an issue with tension. Can anyone please elaborate?

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  • $\begingroup$ " the horse presses the ground in an inclined direction and the horizontal component of the corresponding reaction is the force with which it pulls the cart i.e the tension T" If you mean that the magnitude of the reaction force is equal to the tension if they are moving at constant velocity, then yes, that's true. If you mean that the reaction force is the tension force, then no. $\endgroup$
    – garyp
    Jun 18, 2021 at 14:06
  • $\begingroup$ i'm saying, the Tension force is because of the reaction $\endgroup$
    – rdev
    Jun 18, 2021 at 15:00

2 Answers 2

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This issue eternally confuses beginning students. It certainly confused me when I was a teenager learning mechanics in high school.

The two forces in Newton's third law are on different things ($F_1$ on mass $m_1$, $F_2$ on mass $m_2$ ) and each responds by moving $F_1=m_1a_1$ and $F_2=m_2a_2$ due to the force on it without regard to what the other object is doing. In the abence of any any other forces (not the case with the horse + cart as there is friction) we have $$ 0=F_1+F_2= \frac d{dt}(m_1v_1+m_2v_2), $$
so what does add to zero is the total change of momentum of the two objects. In other words the third law gives you momentum conservation.

In the case of the horse and cart there are additional friction forces that make the horses hooves not slip on the ground, and friction in the cartwheel bearings, and it is the sum of these forces that determines how the combined horse+cart moves.

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You have to focus just on the forces on the cart to decide if it'll move. If the force $T$ on the cart is greater than the friction force resisting the cart, then it'll accelerate in the direction of $T$.

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  • $\begingroup$ ya, i realize that now. thank you. But a different issue just popped up. I edited the question. Will you please re-read and try to answer my query? $\endgroup$
    – rdev
    Jun 18, 2021 at 13:30

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