Bierbank

Relevant wiki: Torque - Dynamical Behavior

Solution edited to reflect the corrected problem.

Consider the torques about the joint on the left (the tipping point).

Clockwise torque due to the table top (concentrating the mass of the table in its COM): $12\cdot 75 = \SI{900}{kgf.cm};$ Counterclockwise torque due to the person in point A: $75 \cdot 17.5 = \SI{1312.5}{kgf.cm}.$ This results in a net torque of $1312.5 - 900 = \SI{412.5}{kgf.cm}$ counterclockwise; we balance this with equal clockwise torque. The arm of the weight on the opposite side is $150 + 17.5 = \SI{167.5}{cm}$ , so that we need a weight of $\frac{412.5}{167.5} = \boxed{\SI{2.46}{kgf}}.$ This is closed to the $\boxed{\text{filled beer mug}}$ option of 2.5 kg.

Alternative solution : find the center of mass of the entire system.

I choose coordinates with $x = 0$ at the center of the plank. The plank's center of mass is at $x = 0$ ; the person's, at $x = -92.5$ cm; and the unknown object's, at $x = +92.5$ cm.

The center of mass of the entire system is the average of centers of mass, $x_{CM} = \frac{12\cdot 0 + 75\cdot -92.5 + m\cdot 92.5}{12 + 75 + m} = \frac{92.5 m - 6937.5}{m + 87}.$ The system will not tip as long as its CM lies between the two supports, i.e. $|x_{CM}| < 75$ . Thus $-75 < \frac{92.5 m - 6937.5}{m + 87} < 75;$ $-75 m - 6525 < 92.5 m - 6937.5 < 75 m + 6525;$ $412.5 - 167.5 m < 0 < 13462.5 - 17.5 m;$ $2.46 < m < 769.2.$ Less than a full beer mug will make the table tip to the left; more than an baby elephant will make it tip to the right; anything in between balances it nicely.

Relevant wiki: Torque - Dynamical Behavior

We consider the torques with respect to the left foot of the bench as the pivot point. The bench's own weight $m$ also provides a contribution. In fact, the wooden panel can be split into two partial masses $\begin{aligned} m_1 &= \frac{d}{l} m = 1.909 \, \text{kg}\\ m_2 &= \frac{l-d}{d} m = 10.091 \, \text{kg}\end{aligned}$ which are respectively located on the right and left of the center of rotation ( $d = 35 \,\text{cm}$ is the overhang). These are represented as point masses, which are located at their respective centers of gravity at $\begin{aligned} x_1 &= - \frac{d}{2} = -0.175\,\text{m} \\ x_2 &= \frac{l-d}{2} = 0.925 \,\text{m}. \end{aligned}$ In addition, the weight $M$ of the person and the compensation mass $M'$ are located at $x_1$ and $x_3 = l - 1.5 \cdot d = 1.675 \, \text{m}$ , respectively. To ensure that the bank does not turn clockwise and tip over, the entire torque must be negative: $\begin{aligned} \sum_{i} T_i &= (m_1 + M) g x_1 + m_2 g x_2 + M' g x_3 < 0 \\ \Rightarrow \quad M' &> -\frac{(m_1 + M) x_1 + m_2 x_2}{x_3} = 2.46 \, \text{kg} \end{aligned}$ Therefore, the filled beer mug is just enough to stabilize the bench.