The question is: Three smooth homogeneous cylinders A, B, C are stacked in a V shaped through as shown in the figure below. Each cylinder weights 100 and has a diameter 5 cm. Determine the forces exerted by the supports on the pipes at contact surfaces.
the angel theta is equal to 30 degrees
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The diameter of the cylinders in this configuration is irrelevant. The forces will be the same for any size of elements with these angles. The equilateral triangle here can be thought of as $2$ right triangles joined with the right angles at the center of the base and the downward force of $100$ is split between these two. Oddly enough, each of the $2$ contact points will feel more than a force of $50$. Here's why. The steeper the angle from horizontal for applied force, the more of that force is needed to transfer part of that force to the upward direction.
The vertical line dividing the equilateral triangle is equal to the slant side (hypotenuse) of right triangle times the cosine of $60$ degrees. Since $\sin{60}=\frac{\sqrt{3}}{2}$, the length of the hypotenuse, and therefore the force vector at the point of contact, will be $\frac{50}{\frac{\sqrt{3}}{2}}=\frac{100}{\sqrt{3}}\approx 57.73502692$.
The forces for the lower contact points on each of the two lower cylinders is $3*57.73502692=173.2050808$.
For the V, the vertical force at each contact surface is half of $300 = 150$.
The normal force is $\frac{150}{\cos(30)} = 173.21$
At the contact points between A and B, and A and C, the vertical force is half of $100 = 50$.
The normal force is $\frac{50}{\cos(30)} = 57.74$ (the normal force here is also at $30$ deg to the vertical).
Between B and C we must react the horizontal component of the $173.21$ force minus the horizontal component of the $57.74$ force. This will be $173.21\sin(30) - 57.74\sin(30) = 86.60 - 28.87 = 57.73$