Three Logs

We are interested in studying what happens to log number 2. We have to consider the two moments with respect to $O$ that affect its stability. We want to find the angle $\theta$ that satisfies: $Mg\,r\sin\theta=F_2\,r\sin30^\circ\quad(1)$ To find $F_2$ consider the vector diagram on log number 3. $F_1$ and $F_2$ are its weight components with respect to the axes that go through the centers of logs 1 and 2 respectively. These two components satisfy: $\begin{aligned} F_1\cos(30^\circ-\theta)+F_2\cos(30^\circ+\theta)&=&Mg\\ F_1\sin(30^\circ-\theta)&=&F_2\sin(30^\circ+\theta) \end{aligned}$ We substitute $F_1$ and solve for $F_2$ : $F_2=\dfrac{Mg\sin(30^\circ-\theta)}{\sin(30^\circ+\theta)\cos(30^\circ-\theta)+\cos(30^\circ+\theta)\sin(30^\circ-\theta)}=\dfrac{Mg(\sin30^\circ\cos\theta-\cos30^\circ\sin\theta)}{\sin60^\circ}=\dfrac{Mg(\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta)}{\frac{\sqrt{3}}{2}}\quad(2)$ where we have used the identity $\sin(\alpha+\beta)=\sin\alpha\cos\beta+cos\alpha\sin\beta$ .

Substituting $(2)$ in $(1)$ and solving for $\theta$ , we obtain the desired result: $\theta=\tan^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)\approx 11^\circ$

Let the top cylinder exerts a force N1 on bottom left and a force N2 on the bottom right cylinder respectively. Then the top cylinder will tend to fall when N2cos(π/3) =Wsin(α) (α is the critical angle of inclination). Applying Lami's theorem to the three force system N1, N2, W we get N2=Wsin(π/6-α)/sin(π/3) Solving the two equations, we get α=ATAN(1/3√3)=10.893 degrees