A metal bar $70 c m$ long and $4.00 k g$ in mass supported on two knife-edges placed $10 c m$ from each end. A $6.00 k g$ load is suspended at $30 c m$ from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)

45 N

and

43 N

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50 N

and

35 N

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55 N

and

43 N

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54 N

and

30 N

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Solution

The correct option is A $45 N$ and $43 N$
Figure 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 as its centre of gravity G. The rod is uniform in cross section and homogenous $;$ hence G is at the centre of the rod $; A B = 70$ cm. $A G = 35$ cm, $A P = 30$ cm, $P G = 5$ cm, $A K_{1} = B K_{2} = 10$ 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 translational equilibrium of the rod, $R_{1} + R_{2} - W_{1} - W = 0$ .... $(i)$
Note $W_{1}$ 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 is G. The moments of $R_{2}$ and $W_{1}$ are anticlockwise $(+ v e)$ , whereas the moment of $R_{1}$ is clockwise $(- v e)$ .
For rotational equilibrium,
$- R_{1} (K_{1} G) + W_{1} (P G) + R_{2} (K_{2} G) = 0$ $(i i)$
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 $(i)$
$R_{1} + R_{2} - 4.00 g - 6.00 g = 0$
or $R_{1} + R_{2} = 10.00 g N$ $(i i i)$
$= 98.00 N$
From $(i i), - 0.25 R_{1} + 0.05 W_{1} + 0.25 R_{2} = 0$
or $R_{1} - R_{2} = 1.2 g$ $N = 11.76 N$ $(i v)$
From $(i i i)$ and $(i v)$ , $R_{1} = 54.88 N$ ,
$R_{2} = 43.12 N$
Thus the reactions of the support are about $55 N$ at $K_{1}$ and $43 N$ at $K_{2}$ .

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