similar to akhil's method but oh well, Norma force=N=mg-Tsin@ frictional force=(mg-Tsin@) at critical pt (mg-Tsin@)=Tcos@ mg=T(cos@/ + sin@) so for mg(max): cos@/ + sin@ is max n v no max value of asin@ + bcos@ = (a2+b2) so mg(max)=T(root(1/^2 +1))
subst T n solve for m..no need diff etc o think..neway it amts to the same thing
EDIT: oops i din no it was the exact same method...pl excuse
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(mg-Tsin@)
(mg-Tsin@)=Tcos@
(a2+b2)
