Question

A block of mass m slides down on rough inclined plane of angle x with the horizontal with an acceleration "a" along the plane the value of reaction is?

Answer 1

@sami-21 @vikrantg4

Answer 2

Normal reaction ?

Answer 3

i think so...

Answer 4

it's mg cos x
:D

Answer 5

but the answer should be

Answer 6

\[m \sqrt{g^2 +a^2 - 2agsinx}\]

Answer 7

omg

Answer 8

Is there any diagram in your book ?

Answer 9

No

Answer 10

Let us assume the block is moving down the plane.

Normal reaction is:
N = mg cos θ

Tangential reaction (friction) is:
T = mg sin θ - ma

Hence total reaction is:

R = √(N² + T²) = \(m \sqrt{g^2 +a^2 - 2ag\sin \theta}\)