A metal bar 70 cm long and 4.00 kg in mass supported on two knifeedges placed 10 cm from each end. A 6.00 kg load 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.)
Example-(6.8)
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Figure 6.26 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 homogeneous; hence G is at the centre of the rod; $AB =70 cm . AG =35 cm , AP$ $=30 cm , PG =5 cm , AK _1= BK _2=10 cm$ 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 translational equilibrium of the rod,
$
R_1+R_2-W_1-W=0
$
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 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\left( K _1 G \right)+W_1( PG )+R_2\left( K _2 G \right)=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)
$
\begin{array}{c}
R_1+R_2-4.00 g-6.00 g=0 \\
\text { or } R_1+R_2=10.00 g N \\
=98.00 N
\end{array}
$
From (ii), $-0.25 R_1+0.05 W_1+0.25 R_2=0$
or $R_1-R_2=1.2 g N =11.76 N$
From (iii) and (iv), $R_1=54.88 N$,
$
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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