Question

A 50kg block is placed on an inclined plane with an angle of 30degree then find the components of the weight (i) perpendicular (ii) parallel to the inclined plane

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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Answer 2

Answer:

fy=75 root 3

And fx=75

Explanation:

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