Physics, asked by singhutkrisht114, 5 months ago

A body of mass m is kept stationary on a rough inclined
plane of angle of inclination
 \alpha
. The magnitude of force acting
on the body by the inclined plane is equal to :
(A) mg
(B) mg sin
 \alpha

(C) mg cos
 \alpha

(D) None​

Answers

Answered by saritadevimdav
0

Answer:

Friction force by the inclined plane is =μmgcosθ=mgsinθ up the inclined plane since the block does not slide.

Normal force perpendicular to inclined plane is mgcosθ.

So, Total force is vector sum of two forces is equal to mg.

N=mgcosθ, ​  

F =mgsinθ

R^2 =N^2 +F^2

⇒R=mg

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