Question

Why do we take a component of mg not Normal in an inclined plane derivation of class 11?

Answer 1

Explanation:

hello mate

we know that the mg force is a vector(because it has both direction and magnitude)

now,

imagine an inclined plane on which ab object is placed at the top.

now,

gravity will act a force downwards.

this force is equal to

$\boxed{\red{\tt force=mg }}$

this has a direction downwards but when you release the object,it will follow the inclined path and the direction of acceleration would not be equal to the direction of gravitational force.

to make the direction of gravitational force equal to the direction of acceleration, we take the component of gravity equal to :

$\boxed{\blue{\tt mgcostheta }}$

hope it helped..

Answer 2

Answer:

we know that the mg force is a vector(because it has both direction and magnitude)

now,

imagine an inclined plane on which ab object is placed at the top.

now,

gravity will act a force downwards.

this force is equal to

this has a direction downwards but when you release the object,it will follow the inclined path and the direction of acceleration would not be equal to the direction of gravitational force.

to make the direction of gravitational force equal to the direction of acceleration, we take the component of gravity equal to :

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Explanation: