To solve this problem, let's use the condition of rotational equilibrium. The plank will be in equilibrium when the sum of the torques about any point is zero.

Given:

Length of the plank: L$=8\mathrm{.}00$mL $=8\mathrm{.}00\backslash ,\backslash$text$\{$m$\}$

Mass of the plank: mplank$=60\mathrm{.}0$kgm$\_\{\backslash$text$\{$plank$\}\}=60\mathrm{.}0\backslash ,\backslash$text$\{$kg$\}$

Position of the left support: $1\mathrm{.}00$m$1\mathrm{.}00\backslash ,\backslash$text$\{$m$\}$ from the left end

Position of the right support: $5\mathrm{.}00$m$5\mathrm{.}00\backslash ,\backslash$text$\{$m$\}$ from the left end

Gravitational acceleration: g$=9\mathrm{.}8$m$/$s$2$g $=9\mathrm{.}8\backslash ,\backslash$text$\{$m$/$s$\}^2$

Let the mass placed at the right end of the plank be maddm$\_\{\backslash$text$\{$add$\}\}.$

Step $1$: Find the weight of the plank

The weight of the plank is given by:

Wplank$=$mplank$$g$=60\mathrm{.}0$kg$9\mathrm{.}8$m$/$s$2=588\mathrm{.}0$NW$\_\{\backslash$text$\{$plank$\}\}=$ m$\_\{\backslash$text$\{$plank$\}\}\backslash$times g $=60\mathrm{.}0\backslash ,\backslash$text$\{$kg$\}\backslash$times $9\mathrm{.}8\backslash ,\backslash$text$\{$m$/$s$\}^2=588\mathrm{.}0\backslash ,\backslash$text$\{$N$\}$

Step $2$: Calculate the torques

We'll take torques about the left support point $($located at x$=1\mathrm{.}00$mx $=1\mathrm{.}00\backslash ,\backslash$text$\{$m$\}).$

Torque due to the plank's weight: The weight of the plank acts at its center of mass, which is at $4\mathrm{.}00$m$4\mathrm{.}00\backslash ,\backslash$text$\{$m$\}($half the length of the plank$).$ The distance from the left support to the center of mass is $4\mathrm{.}001\mathrm{.}00=3\mathrm{.}00$m$4\mathrm{.}00-1\mathrm{.}00=3\mathrm{.}00\backslash ,\backslash$text$\{$m$\}.$

plank$=$Wplank$3\mathrm{.}00$m$=588\mathrm{.}0$N$3\mathrm{.}00$m$=1764\mathrm{.}0$N$$m$($counterclockwise$)\backslash$tau$\_\{\backslash$text$\{$plank$\}\}=$ W$\_\{\backslash$text$\{$plank$\}\}\backslash$times $3\mathrm{.}00\backslash ,\backslash$text$\{$m$\}=588\mathrm{.}0\backslash ,\backslash$text$\{$N$\}\backslash$times $3\mathrm{.}00\backslash ,\backslash$text$\{$m$\}=1764\mathrm{.}0\backslash ,\backslash$text$\{$N$\}\backslash$cdot $\backslash$text$\{$m$\}\backslash$quad $(\backslash$text$\{$counterclockwise$\})$

Torque due to the additional mass: The mass maddm$\_\{\backslash$text$\{$add$\}\}$ is placed at the right$-$hand end of the plank, which is $8\mathrm{.}005\mathrm{.}00=3\mathrm{.}00$m$8\mathrm{.}00-5\mathrm{.}00=3\mathrm{.}00\backslash ,\backslash$text$\{$m$\}$ away from the right support. The distance from the left support is $8\mathrm{.}00$m$1\mathrm{.}00$m$=7\mathrm{.}00$m$8\mathrm{.}00\backslash ,\backslash$text$\{$m$\}-1\mathrm{.}00\backslash ,\backslash$text$\{$m$\}=7\mathrm{.}00\backslash ,\backslash$text$\{$m$\}.$

add$=$madd$$g$7\mathrm{.}00$m$\backslash$tau$\_\{\backslash$text$\{$add$\}\}=$ m$\_\{\backslash$text$\{$add$\}\}\backslash$times g $\backslash$times $7\mathrm{.}00\backslash ,\backslash$text$\{$m$\}$

This torque will act clockwise because the mass is located to the right of the support.

Step $3$: Set up the equilibrium condition

For the system to be in equilibrium, the sum of the torques must be zero:

$$$=0\backslash$sum $\backslash$tau $=0$plank$=$add$\backslash$tau$\_\{\backslash$text$\{$plank$\}\}=\backslash$tau$\_\{\backslash$text$\{$add$\}\}1764\mathrm{.}0$N$$m$=$madd$9\mathrm{.}8$m$/$s$27\mathrm{.}00$m$1764\mathrm{.}0\backslash ,\backslash$text$\{$N$\}\backslash$cdot $\backslash$text$\{$m$\}=$ m$\_\{\backslash$text$\{$add$\}\}\backslash$times $9\mathrm{.}8\backslash ,\backslash$text$\{$m$/$s$\}^2\backslash$times $7\mathrm{.}00\backslash ,\backslash$text$\{$m$\}$

Step $4$: Solve for maddm$\_\{\backslash$text$\{$add$\}\}$

$1764\mathrm{.}0=$madd$9\mathrm{.}87\mathrm{.}001764\mathrm{.}0=$ m$\_\{\backslash$text$\{$add$\}\}\backslash$times $9\mathrm{.}8\backslash$times $7\mathrm{.}00$madd$=1764\mathrm{.}09\mathrm{.}87\mathrm{.}00=1764\mathrm{.}068\mathrm{.}625\mathrm{.}7$kgm$\_\{\backslash$text$\{$add$\}\}=\backslash$frac$\{1764\mathrm{.}0\}\{9\mathrm{.}8\backslash$times $7\mathrm{.}00\}=\backslash$frac$\{1764\mathrm{.}0\}\{68\mathrm{.}6\}\backslash$approx $25\mathrm{.}7\backslash ,\backslash$text$\{$kg$\}$

Final Answer:

The maximum additional mass that can be placed on the right$-$hand end of the plank is approximately $25\mathrm{.}7$kg$\backslash$boxed$\{25\mathrm{.}7\backslash ,\backslash$text$\{$kg$\}\}.$