New solution

Tete edited
revision #20612 newer →
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+### Statement
+
+$2.7.28.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $2.7.28$ |490x380, 31%](../../img/2.7.28/Savechenko.png)
+
+Since the moment of inertia and the clockwise moment with respect to the center of mass are $ml^2/12$ and $Na/2$, respectivly, the angular acceleration is
+
+\[\frac{Na/2}{ml^2/12}=\frac{6Na}{ml^2}.\]
+
+Similarly, since the moment of inertia and the clockwise moment with respect to the remaining support are $ml^2/12+m(a/2)^2$ and $mga/2$, respectively, the angular acceleration is
+
+\[\frac{mga/2}{ml^2/12+m(a/2)^2}=\frac{6ga}{l^2+3a^)}.\]
+
+Equating the two expressions for the angular acceleration, we have
+
+\[\frac{6Na}{ml^2}=\frac{6ga}{l^2+3a^2}\qquad\Rightarrow\qquad N=\frac{mgl^2}{l^2+3a^2}.\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]