If action is always equal to the reaction, Explain how a horse can pull a cart.
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Answered by
25
according to Newton' s 3rd law , we know, action and reaction are of equal magnitude and directed in opposite directions.
now , in case of horse and cart
===========================
horse pull the cart forward direction , but how cart reacted to this force ?
actually , here frictional force do important work .
we know, frictional force oppose the relative motion of particle or body .
also we know, frictional force is adjusting but limiting .
Let force applied by horse = F
and let net frictinal force appear all feet of horse and cart's lowest part f
then,
Fnet = F -f { frictional force adjusting but limitng }
now from Newton's 2nd law ,
Fnet = ma
so, ma = F -f
here we see a is appear . hence, cart is moved forward . hence, we can say that here cart applied reation force indirectly (e.g frictional force )
now , in case of horse and cart
===========================
horse pull the cart forward direction , but how cart reacted to this force ?
actually , here frictional force do important work .
we know, frictional force oppose the relative motion of particle or body .
also we know, frictional force is adjusting but limiting .
Let force applied by horse = F
and let net frictinal force appear all feet of horse and cart's lowest part f
then,
Fnet = F -f { frictional force adjusting but limitng }
now from Newton's 2nd law ,
Fnet = ma
so, ma = F -f
here we see a is appear . hence, cart is moved forward . hence, we can say that here cart applied reation force indirectly (e.g frictional force )
abhi178:
how is this i don't know, but i use which is possibility
Answered by
19
Action force is always equal to reaction force. However, action force is exerted by horse on the cart, when horse pulls the cart. But reaction force (equal and in opposite direction) is exerted by the cart on the horse.
If you consider forces on horse, horse pushes the ground backwards. The push force exerted by the ground (friction force) (minus the reaction force exerted by the cart) pushes the horse forward. So friction from ground is more than action force.
If you see the forces on the cart, the action force of horse (minus the friction force on the cart) pushes the cart forward.
See the diagram. See the free body diagrams on horse, ground under the feet of horse, cart and the ground under the cart.
F1 is action force by horse on cart. F2 (= F1) is reaction force of cart on horse.
F3 is action force of push by horse on ground. F4 (= F3) is reaction force of ground on horse.
F5 is the action force of push of the wheel on ground. F6 (= F5) is the reaction force of ground on cart.
F4 - F2 = m1 * a for Horse.
F1 - F5 = m2 * a for Cart
a = acceleration is same for both horse and cart.
If you consider forces on horse, horse pushes the ground backwards. The push force exerted by the ground (friction force) (minus the reaction force exerted by the cart) pushes the horse forward. So friction from ground is more than action force.
If you see the forces on the cart, the action force of horse (minus the friction force on the cart) pushes the cart forward.
See the diagram. See the free body diagrams on horse, ground under the feet of horse, cart and the ground under the cart.
F1 is action force by horse on cart. F2 (= F1) is reaction force of cart on horse.
F3 is action force of push by horse on ground. F4 (= F3) is reaction force of ground on horse.
F5 is the action force of push of the wheel on ground. F6 (= F5) is the reaction force of ground on cart.
F4 - F2 = m1 * a for Horse.
F1 - F5 = m2 * a for Cart
a = acceleration is same for both horse and cart.
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