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A uniform solid rod of mass 40kg and len...

A uniform solid rod of mass 40kg and length 10m rests against a vertical smooth wall making an angle of `30^(@)` with the vertical. Find the force of friction (f) and normal reaction (R ) exerted by the groundd on the rod.

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For rotatory equilbrium,

`Fbar(AB)=mgbar(CD)`
AC=10m,m=40kg
`sin60^(@)=(AB)/(AC)" ":.bar(AB)=bar(AC) sin60^(@)=10xx0.866=8.66m`
`sin30^(@)=(BC)/(AC)" ":.BC=10xx0.5=5m`
CD=`(BC)/2=2.5m" ":.F(8.66)=40xx9.8xx2.5`
i.e `F=(40xx9.8xx2.5)/(8.66)=113.16N` Force of friction=113.16N
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