Tripod

Consider the forces on system , there are 3 upward normal reactions ( $N$ ) each from the ground on legs of tripod , and the weight of system , Balancing force on the system, we get :

$N = \frac{3mg + Mg}{3}$ .

Now , consider $B,C$ and $D$ be the vertices of triangle made by massless strings and P as the centroid of triangle. Let $\theta$ be the angle made by a rod with vertical,

In $\triangle APB,$ $sin\theta = \frac{PB}{AB} = \frac{\frac{b}{\sqrt{3}}}{\frac{a}{2}}$ = $\frac{1}{2\sqrt{3}}$ , $\Rightarrow tan\theta = \frac{1}{\sqrt{11}}$ .

Now , firstly net force of tension acting on B = $2Tcos\frac{\pi}{6} = T\sqrt{3}$

Balance torque on a rod about A to get,

$Nasin\theta = mg\frac{a}{2}sin\theta + T\sqrt{3}\frac{a}{2}cos\theta$

$\Rightarrow T = \frac{2N - mg}{\sqrt{3}} tan\theta = \frac{2Mg + 3mg}{3\sqrt{33}} = \fbox{2.5}$

Lets set up a coordinate system with origin at one of the legs of tripod.

Tripod

G is the centroid of triangle OBC.

The coordinates can be easily found out, I am skipping that task.

Note the CM of the complete system lies on the blue line, I haven't shown the exact position of it as it is not required.

There are some unknown forces at the pivot but since they cancel, we don't have to care about including them in the force balance equation.

I have only shown the tension force acting on the leg OA.

Since the system is at rest or equilibrium, the net force is zero. Summing the forces in the y-direction,

$\displaystyle N_1+N_2+N_3=(M+3m)g=3.4g$

From the symmetry, we can conclude that $N_1=N_2=N_3=N$

Hence, $N=3.4g/3$ .

Next we balance the moments of forces on leg OA about point A. The point A is chosen because the moment of the unknown forces mentioned before is zero.

The force acting at point D is: $\displaystyle \vec{F_D} = T\hat{i}+\frac{T}{2}\hat{i}+\frac{T\sqrt{3}}{2}\hat{k}-mg\hat{j}$

$\displaystyle \Rightarrow \vec{F_D}=\frac{3T}{2}\hat{i}-mg\hat{j}++\frac{T\sqrt{3}}{2}\hat{k}$

The force acting on leg OA at origin is: $\vec{F_O}=N\hat{j}$ .

The net moment of the forces about A is: $\vec{AD} \times \vec{F_D}+\vec{AO}\times \vec{F_O} =0$

We have

$\displaystyle \vec{AD}=\vec{OD}-\vec{OA}=\frac{-b}{2}\hat{i}+(-b)\sqrt{\frac{11}{3}}\hat{j}+\frac{-b}{2\sqrt{3}}\hat{k}$

and

$\displaystyle \vec{AO}=-b\hat{i}+(-2b)\sqrt{\frac{11}{3}}\hat{j}+\frac{-b}{\sqrt{3}}\hat{k}$

Evaluating the cross products is trivial. Equate any of the component of the net moment to zero to get the value of T.

For instance, equating the x component to zero, you end up with

$\displaystyle \frac{N}{\sqrt{3}}=\frac{T\sqrt{11}}{2}+\frac{mg}{2\sqrt{3}}$

Solving for T, $\fbox{T=2.50208 N}$ .

I do understand that its a quite long solution. I am interested to know how other members approached this.

I took the utmost care to prevent my solution from errors. I am sorry if errors have prevailed.

Consider the forces that act upon one rod. These are the weight of the camera $Mg$ , the weight of the rod itself $mg$ , the normal force on the point that touches the ground $M/3 + mg$ (this is true since the projections of forces on the vertical axis are equal as the system is in equilibrium) and the resultant force $F$ from tensions $T$ of two strings.

Since the system is in equilibrium, the sum of moments of forces must equal zero. Let an angle that the rod makes with the ground be $\alpha$ and let the angle $\frac{\pi}{2} - \alpha = \beta$ . Then the moments of forces about point $A$ are the following: $$M_{Mg} = 0\text{,}$$

\begin{array} \\ M {mg} &= \int 0^a a' g \sin{\beta} dm \\ &= \int_0^a a' g \sin{\beta} \rho da' \\ &= \frac{mg a \sin{\beta}}{2}, \\ \end{array}

$$M_F = \frac{Fa\sin{\alpha}}{2},$$

$$M_N = Na \sin{\beta} = (\frac{M}{3} + m)ga \sin{\beta}.$$

Equating the sum of moments with $0$ , making some algebraic manipulations and noting that $\sin{\beta} = \cos{\alpha}$ we get that $$F = g (\frac{2M}{3} + m) \cot{\alpha} \text{ } (*).$$

Now express $F$ in terms of $T$ . It can be done using the Law of Cosines and considering a rhombus with side length $T$ , angle of $60^{\circ}$ and greater diagonal $F$ . It follows that $F = T \sqrt{2 (1 + \cos{60^{\circ}})} = T\sqrt{3})$ .

Substituting this expression for $F$ into $(*)$ we get $$F = \frac{1}{\sqrt{3}} g (\frac{2M}{3} + m) \cot{\alpha}$$

Now all that's left is to find $\alpha$ . First, consider the side of our pyramid. Once again, use Law of Cosines to get that the plane top angle is $\cos^{-1}{\frac{7}{8}}$ and that the side base is of length $\frac{a}{2}$ . Then, consider the base and use Law of Cosines to find that the distance from the height of the pyramid to one of the base vertices is $\frac{a}{\sqrt{12}}$ . Use Pythagoras theorem to find that the height of the pyramid is $\frac{a}{2}\sqrt{\frac{11}{3}}$ . Hence, $\cot{\alpha} = \frac{1}{\sqrt{11}}$ .

Plugging the numbers in we get $$T = \frac{1}{\sqrt{11 * 3}} \cdot 9.8 \cdot (0.8 + \frac{2}{3}) = \boxed{2.5 \text{ (N)}}$$.