Question

20. A horizontal force F acts on a block of mass m placed on a fixed inclined plane, to hold in position,as shown in figure. The normal reaction N on the block is
(1) mgsino + Fcoso
(3) mgcose + Fsino
(2) mgsino – Fcoso
(4) mgcoso – Fsino​

Answer 1

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❄Actually welcome to the Concept of the Mechanics

❄Basically here, after resolving the Problem and Creating a Free body Diagram,

❄We get as

❄N = mgcos© - Fsin©

Answer 2

$\bold {\huge {Answer :-}}$

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$\normalsize\bold{Let\:1^{st}\:draw\:FBD\:}$

➛see attachment

Now,

Along Inclination

→Since block is fixed therefore no acceleration along the plane but the components of $\normalsize\bold{mg}$will act.

Which will balance the force component along the plane

$\normalsize\bold{\therefore\:mgsin\:\theta\:= mgcos\:\theta}$

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Along perpendicular to the inlined

Downward Perpendicular to the plane will act to force 2 force by the block . One is component of mg other is component of F

To balance Force = 0 , Plane will act equal force to the block known as Normal Force (N)

That is

$\normalsize\bold{N\:=mgcos\:\theta\:+ Fsin\:\theta}$

Hence, the normal force is $\normalsize\bold{mgcos\:\theta\:+ Fsin\:\theta}$

→Option (3)