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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). |
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Answer» Solution :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` |
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