Question

The figure shows a horizontal force 7 acting on a
block of mass Mon an inclined plane (angle 0). What
is the normal reaction on the block ?
(1) mgsino + Fcoso
(3) mgcoso - Fsino
(2) mgsino - Fcoso
(4) mgcoso + Fsino​

Answer 1

the weight mg=50×10=500N

now as it is on inclined plane it can be resolved into 2 components

which are parallel and perpendicular to the plane.

now let us see first component perpendicular to the plane.

we know that force which acts perpendicular to plane is normal reaction force.

here actually we will be finding normal reaction.

this will be equal to mgcos30 as the angle made by mg and perpendicular of plane is 30. so by vector resolution it will make cos30 along perpendicular. so our answer is mgcos30.

now m=50 let's take g=10 cos30=√3/2

so mgcos30= 50×10×√3/2= 250√3

now let us see the parallel case.

again as angle made perpendicular and mg is 30. or the angle made by mg and inclined plane is 60

again by vector resolution we get force along inclined plane=mgsin30 or mgcos60

now putting values

m=50 g=10 sin30=1/2

force=50×10×1/2=250N

so force perpendicular is 250√3N

and force parallel is 250N.

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