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.

A uniform solid rod of mass 40kg and length 10m rests against a vertic

1.	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.
Answer» SOLUTION :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