- #1
Soumalya
- 183
- 2
Hi,
I am facing some issues while dwelling upon the application of Newton's third law of motion to the famous "Horse-Cart" Problem.Anybody who is clueless about what I am talking about please follow the link:
https://www.lhup.edu/~dsimanek/physics/horsecart.htm
As one can see the whole system has been sub divided into two sub systems.For the sub system comprising the horse and the ground we have two forces acting on the horse:
1. The horse pushes the ground with it hooves and so does the ground equally to the horse in an opposite direction.
2.The cart is pulled by the horse (say with the help of a cable connecting the horse and the cart) and so the cart pulls back the horse with an equal and opposite reaction force.This would the tension in the cable.
Referring to the link above from the free body diagram of the horse the first force is D and the second force is C.
Again for the sub system comprising of the cart and the ground we have two forces acting on the cart:
1.The friction between the wheels and the ground in a direction opposite to impending motion or motion.
2.The force on the cart by the horse i.e, the tension in the cable connecting the horse and the cart.
From the free body diagram of the cart the first force is A and the second force is B.
Now, the force applied by the ground on the horse should be equal to the force applied by the horse on the cart as both horse and cart are constrained to each other.(I suppose I am correct here?)
So, D=B.
Again according to the Newton's third law of motion, B=-C.
Thus we should have D=-C.
Apparently the analysis discards any possibility of motion for the horse cart system!
Could anyone outline where did I went wrong?
I am facing some issues while dwelling upon the application of Newton's third law of motion to the famous "Horse-Cart" Problem.Anybody who is clueless about what I am talking about please follow the link:
https://www.lhup.edu/~dsimanek/physics/horsecart.htm
As one can see the whole system has been sub divided into two sub systems.For the sub system comprising the horse and the ground we have two forces acting on the horse:
1. The horse pushes the ground with it hooves and so does the ground equally to the horse in an opposite direction.
2.The cart is pulled by the horse (say with the help of a cable connecting the horse and the cart) and so the cart pulls back the horse with an equal and opposite reaction force.This would the tension in the cable.
Referring to the link above from the free body diagram of the horse the first force is D and the second force is C.
Again for the sub system comprising of the cart and the ground we have two forces acting on the cart:
1.The friction between the wheels and the ground in a direction opposite to impending motion or motion.
2.The force on the cart by the horse i.e, the tension in the cable connecting the horse and the cart.
From the free body diagram of the cart the first force is A and the second force is B.
Now, the force applied by the ground on the horse should be equal to the force applied by the horse on the cart as both horse and cart are constrained to each other.(I suppose I am correct here?)
So, D=B.
Again according to the Newton's third law of motion, B=-C.
Thus we should have D=-C.
Apparently the analysis discards any possibility of motion for the horse cart system!
Could anyone outline where did I went wrong?
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