To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
[tex]F = mg[/tex]
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
[tex]\sum \tau = 0[/tex]
Regarding the forces we have,
[tex]3Mg-1mg=0[/tex]
Re-arrange to find M,
[tex]M = \frac{m}{3}[/tex]
[tex]M = \frac{50}{3}[/tex]
[tex]M = 16.67Kg[/tex]
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
