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A uniform cube of mass m & side a is rsting on a rough 45 degree inclined surface. The distance of the point of application of normal reaction measured from the lower edge of the cube is:
a) zero
b) a/3
c) a/√2
d) a/4
hi,
as normal rxn passes through centre & perpendicular to lower & upper faces & gap between centre & lower edge is
a/21/2
so (c) is correct.
no!!! ans is (a)
sorry for wrong ans.
now we will approach it as a case of toppling
MgsinΘ is the component of weight which acts on centre at a distance a/2 from lower edge. so its torque will be
Τ1= MgsinΘ*a/2(.)
& normal will nullify this effect so
T2 = MgcosΘ*r(×)
equating both
r=a/2 (from centre)
as normal passes through lower face
hence normal will pass through edge.
regards
anand
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