Answer
5503.68 N
Work Step by Step
When the system is in equilibrium, the buoyant force is equal to the total weight of the ice & bear.
$F_{B}=m_{i}g+m_{b}g$
According to principle of Archimedes, we can write,
$m_{w}g=m_{i}g+m_{b}g-(1)$
We know that, $mass = (Density)\times(volume)=\rho V$, Therefore (1)=>
$\rho_{w}V_{w}=\rho_{i}V_{i}+m_{b}$
$m_{b}=\rho_{w}V_{w}-\rho_{i}V_{i}-(2)$
When the heaviest possible bear is on the ice, the ice is fully submerged inside the water. Therefore,
$V_{w}=V_{i}$
(2)=>
$m_{b}=\rho_{w}V_{i}-\rho_{i}V_{i}$
$W_{bmax}=(\rho_{w}-\rho_{i})V_{i}g$ ; Let's plug known values into this equation.
$W_{bmax}=(1025\space kg/m^{3}-917\space kg/m^{3})(5.2\space m^{3})(9.8\space m/s^{2})=5503.68\space N$
