If a horse pulls the carriage, the carriage should pull the horse with equal force (Newton's third law), then how does the whole system move? I drew the free body diagram but not yet figured what makes the whole thing move. Please give me an explanation.
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The system horse+carriage is in equilibrium. However the horse's feet (burning calories) will generate a force on the floor. By third law the floor will generate a force on the system horse+carriage.
The most interesting thing is that the floor will move backwards as well! This is. If the horse+carriage are on the Earth, when it moves due to calories being burned in the horse's muscles, then Earth also moves to the other direction. However if you compare masses, Earth will move practically nothing.
Vladimir Vargas
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The statement that action and reaction act on two different bodies is not sufficient to explain the problem in question. In this case the cart is harnessed to the horse. Thus, they are linked together and travel as a single connected system. The forces of interaction between the horse and the cart are applied to different parts of the same system. In the motion of the system as a whole, these forces can be regarded as mutually counterbalancing forces. Thus the interaction of horse & cart alone cannot impart any acceleration to the system . Or in other words, some outside agency must participate in the problem in addition to the horse & the cart.
In order to explain the forward motion of the horse-cart let us first consider the forces acting on the horse & the cart.
Two forces act on the cart: i) Tension $T$ on the rope connected to it. ii) Friction $F$ on its wheels.
Again, the two forces acting on the horse are: i) Ground reaction $R$ whose horizontal component pushes the horse forward. Considering the two forces acting on the cart, we may say that if $T > F$ , it will accelerate. If $m_1$ is its mass & $f$ is its acceleration, then by 2nd law of motion, we may write $$T - F = m_1 \cdot{f}$$ Again,considering the forces on the horse, we obtain $$H - T = m_2 \cdot{f}$$ where $m_2$ is the mass of the horse & $H$ is the horizontal component of ground reaction. Eliminating $T$ from both the equations , we get $$H - F = (m_1 + m_2)\cdot{f} \implies f = \frac{H - F}{m_1 + m_2}$$ . This is the common acceleration of the system. When $H = F$ , the system under consideration will move with constant velocity. [Note: Though $T$ is absent in the equation of the common acceleration of the system, it is necessary to overcome friction.]