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

Figure 7.26 shows the rod AB, the positions of the knife edges `K_(1) " and " K_(2)`, the centre of gravity of the rod at G and the suspended load at P.
Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homegeneous , hence G is at the centre of the rod, AB=70 cm. AG=35 cm, AP=30 cm , PG=5, `AK_(1)=BK_(2)=10 cm " and " K_(1)G=K_(2)G=25 cm`. Also, W=weight of the rod =4.00 kg and `W_(1)`= suspended load =6.00 kg, `R_(1) " and " R_(2)` are the normal reactions of the support at the knife edges.
For transiational equilibrium of the rod, `R_(1)+R_(2)-W_(1)-W=0`
Not `W_(1)` and W act vertically down and `R_(1) " and R_(2)` act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments of the forces. A convenient point to take moments of the forces. A convenient point to take moments about is G. The moments of `R_(2) " and " W_(1)` are anticlockwise `(+ve)`,whereas the moment of `R_(1)` is clockwise `(-ve)`.
For rotational equilibrium.
`-R_(1)(K_(1)G)+W_(1))PG)+R_(2)(K_(2)G)=0`
It is given that W=4.00 g N and `W_(1)=6.00 g` N, where g=acceleration due to gravity. We take `g=9.8 m//s^(2)`.
With numerical values inserted, from (i)
`R_(1)+R_(2)-4.00g-6.00g=0`
or `R_(1)+R_(2)=10.00 g N`
=98.00 N
From (ii), `-0.25R_(1)+0.05 R_(2)=0`
From (iii) and (iv) , `R_(1)=54.88 N`.
`R_(2)=43.12 N`
Thus the reactions of the support are about 55N at `K_(1) " and " 43 N " at" K_(2)`.