A metal bar `70 cm` long and `4.00 kg` in mass is supported on two knife edges placed `10 cm` from each end. A `6.00 kg` weight is suspended at `30 cm` from one end. Find the reactions at the knife edges. Assume the bar to be of uniform cross-section and homogeneous.

Explanation

Here is the centre of gravity of the rod where the weight W of the rod acts. Let P be the point, 30 cm from end A from where W. is suspended. `K_(1)andK_(2)` are the knife edges and `R_(1)` and `R_(2)` are the reactions of the support of edges respectively.
AB=70 cm, AP=30 cm
AO = 35 cm,
PO = 5 cm,
`AK_(1)=BK_(2)=10cm`
`K_(1)O=K_(2)O=25cm` `W=4.00kgwt=4gN`
`W.=6kgwt=6gN`
For translational equilibrium of the rod,
`R_(1)+R_(2)=W+W.=4g+6g=10g`
`impliesR_(1)+R_(2)=10xx9.8=98N" "......(i)`
For rotational equilibrium,
Clockwise moment = Anticlockwise moment
`R_(1)xxK_(1)O=W.xxPO+R_(2)xxK_(2)O`
`impliesR_(1)xx0.25=W.xx0.05+R_(2)xx0.25`
`implies0.25(R_(1)-R_(2))=0.05W`
`(R_(1)-R_(2))=0.2xx6g=0.2xx6xx9.8`
`=11.76N" ".......(ii)`
Solving (i) and (ii), we get
`R_(1)=54.88NandR_(2)=43.12N`