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

Explanation

The rod AB, the positions of the knife edges `K_(1)` and `K_(2)` is shown in figure. The centre of gravity of the rod at G and the suspended load is P.
The weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous, hence G is at the centre of the rod,
`AB=70cm`
`therefore AG=BG=35cm`
`PG=5cm therefore AP=30cm`
`AK_(1)=BK_(1)=10cm`
`K_(1)G=K_(2)G=25cmandg=9.8ms^(-2)`
and weight of rod `W=4kgxx9.8=39.2N` suspended load at P, `W_(1)=6kgxx9.8=58.8N`
`R_(1)andR_(2)` are the normal reaction of the support at the knife edge `K_(1)andK_(2)` respectively.
For the translation equilibrium of the rod
`R_(1)+R_(2)-W_(1)-W=0" "...(1)`
`therefore R_(1)+R_(2)-58.8-39.2=0`
`therefore R_(1)+R_(2)=98" "...(2)`
`W_(1)` and W act vertically down and `R_(1)andR_(2)` act vertically up.
For rotational equilibrium
`therefore -R_(1)(K_(1)G)+W_(1)(PG)+R_(2)(K_(2)G)=0...(3)`
`therefore -R_(1)(0.25)+58.8(0.05)+R_(2)(0.25)=0`
`therefore 0.25R_(1)-0.25R_(2)=2.94`
`therefore 0.25(R_(1)-R_(2))=2.94`
`therefore R_(1)-R_(2)=(2.94)/(0.25)`
`therefore R_(1)-R_(2)=11.76N" "...(4)`
Sum of eqn. (2) and (4),
`2R_(1)=109.76`
`therefore R_(1)=54.88NorR_(1)~~55N`
In eqn. `R_(1)+R_(2)=98,R_(1)=54.88`
`R_(2)=98-54.88`
`therefore R_(2)=43.12N or R_(2)=43N`