Question

A rectangular block of ice with dimensions 2 m by 2 m by 0.3 m floats on water. A person weighing 835 N wants to stand on the ice. Would the ice sink below the surface of the water?

Answer 1

If $F_B$ (Buoyant Force) $\ge$ (m+M) g

Then the ice will float.

$F_B=(\rho_wV_{ice}) g$

$=(1000×2×2×0. 2)9.8$

$=7840N$

$mg+Mg=980+(\rho_{ice}V_{ice} g)$

$=(980+(916.7×2×2 ×0. 2×9.8)$

$=980+7186.93$

$=8166.93N \gt F_B$

The ice will sink since $F_B$ is not greater than $(m+M) g$