Answer:
(b,
d) Let
body moves up the plane, then
\[{{m}_{1}}a=T{{m}_{1}}\,\,g\text{
}\sin \theta \mu R\]
\[{{m}_{1}}a=T{{m}_{1}}g\text{
}\sin \theta \mu \text{ }{{m}_{1}}g\text{ }\cos \theta \] ?. (i)
Also
\[{{m}_{2}}a={{m}_{2}}gT\] ? (ii)
\[\therefore
\] \[({{m}_{1}}+\,{{m}_{2}})\,a=\,{{m}_{2}}\,g\,-{{m}_{1}}\]\[(g\,\sin
\,\theta +\,\mu \,g\,\cos \theta )\]
\[=\,[{{m}_{2}}-{{m}_{1}}\,(\sin
\theta \,+\,\mu \,\cos \theta )]g\]
\[a=\,\frac{[{{m}_{2}}-{{m}_{1}}\,(\sin
\,\theta \,+\mu \,\cos \theta )]\,g}{({{m}_{1}}+\,{{m}_{2}})}\]
The
body will move up if \[{{m}_{2}}>{{m}_{1}}\]
\[(\sin
\,\theta \,\,+\,\mu \,\cos \theta )\]
Let
body moves down the plane, then
\[{{m}_{1}}a=\,{{m}_{1}}g\,\,\sin
\,\theta -\,T-\mu R\]
\[={{m}_{1}}\,g\,\sin
\,\theta \,-T-\,\mu {{m}_{1}}\,g\,\cos \theta \]
Also
\[{{m}_{2}}\,a=\,T-\,{{m}_{2}}g\]
\[\therefore
\]\[({{m}_{1}}+\,{{m}_{2}})\,a\]\[=\,[{{m}_{1}}\,(\sin \,\theta \,-\mu \,\cos
\theta )-{{m}_{2}}]g\]
or \[a=\,\frac{[{{m}_{2}}\,-{{m}_{1}}]\,(\sin
\,\theta +\,\mu \,\cos \theta )-\,{{m}_{2}}]\,g}{({{m}_{1}}+\,{{m}_{2}})}\]
Thus,
body will move down the plane, if
\[{{m}_{2}}<{{m}_{1}}(\sin
\,\theta -\,\mu \,\,\cos \theta )\]
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