Question

Three identical rigid circular cylinders A, B and C are arranged on smooth inclined surfaces as shown in figure. Find the least value of theta that prevent the arrangement from collapse.The link of the image is here- http://prntscr.com/1tfxk3

Answer 1

$mg - 2N\cos \beta = 0$ - for the first cylinder

For the second cylinder :

$\mathrm{OY}:mg + N\cos \beta -N_0\cos \theta = 0$

$\mathrm{OX}:N_{1} + N\sin \beta -N_{0}\sin \theta = 0$

$$\beta = 3 0 ^ {0}$$

When the system starts to destroy $N_{1} = 0$

From the first equation: $\frac{mg}{2\cos\beta} = N$

From the second equation: $N_0 = \frac{3mg}{2\cos\theta}$

So, in the third equation we have:

$$\begin{array}{l} \frac {m g}{2} t g \beta - \frac {3 m g}{2} t g \theta = 0 \\ t g \beta = t g 3 0 ^ {0} = \frac {1}{\sqrt {3}} \\ = > t g \theta = \frac {1}{3 \sqrt {3}} \\ \theta = \operatorname {arctg} \left(\frac {1}{3 \sqrt {3}}\right) \quad \longrightarrow \quad \text {Answer} \\ \end{array}$$