Question

horse and cart derivation

Answer 1

According to Newton's Third law of motion, every action, there is always an equal (in magnitude) and opposite (in direction) reaction.

In case of horse cart, The horse pushes the ground with it's foot in the backward direction by pressing the ground. As a result of this force of action ( i.e. backward push ), the ground pushes the horse in forward direction

According to Newton's third law of motion , The Ground pushes the horse in forward direction. Thus, the horse moves in a forward direction . As the cart is attached to the horse , the cart too moves in the forward direction.

Hence, The horse pulls the cart .

( See the Diagram )
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Let FH = Force exerted by Horse on Cart.

FC = Force exerted by cart on Horse.

FH & FC form Action-Reaction pair of forces.

FG = Force of ground on the horse.

In Addition, there is a Frictional Force, " f " that acts on the cart.

The Cart accelerates Because,
F = FH - f , due to this unbalanced force acting on it.

The Cart begins to move as soon as
FH > f

Let F' = net force acting on the horse
F' = FG - FC

The horse will accelerate due to this unbalanced force. Although, the cart is pulling the horse backward, the ground exerts a forward force on the horse. Hence, it accelerates.

The Action-Reaction pair, FC and FH do not cancel out, as they act on two different bodies.

Answer 2

H=force of horse

F=force of friction

M=mass of horse

G=acc due to gravity

H.m=mass of cart

T=tention of string

H-T=Ma--(1)

Motion of cart

T-F=ma--(2)

Adding 1 and 2

H-T=Ma/-f+t=ma(by cutting them we get)

H-F=a(M+m)

H-F/M+m=a(from this we get acceleration

For finding tention

Divide equation1&2

H-T/T-F=Ma/ma(by cutting we get)

m(H-T)=M(T-F)

mh-mt=MT-Mf

Mh+Mf=MT+my

Mh+Mf=t(M+m)

mh+Mf/M+m=T