# A diver in water picks up a lead cube of side length `10 cm` . How much force is needed to lift the cube? The density of lead is approximately `11 g/(cm^3)` .

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There are three forces acting on the lead cube:

gravitational force with the magnitude `Mg` downwards,

buoyant (Archimedes') force `mg` upwards,

lifting force of unknown magnitude `F` upwards.

Here `g` is the gravity acceleration, M is the mass of the cube and m is the mass of water in the volume of the cube. Note that `M = rho_l V,` `m = rho_w V,` `V = a^3,` where `a` is the side length, `V` is the volume of the cube, `rho_l` is the density of lead and `rho_w approx1 g/(cm)^3` is the density of water.

If we find `F_0` such that these forces will be balanced, then any lifting force magnitude greater than `F_0` value will be suitable.

Considering the directions, the equation becomes

`F_0 +rho_w a^3 g =rho_l a^3 g,` or

`F_0 =rho_l a^3 g -rho_w a^3 g = (rho_l -rho_w) a^3 g.`

We need to divide this by `1000` to obtain kilograms from grams. The numerical result is about `(11 - 1)*10^3*10/1000 = 100 (N).`

Thus the answer is: at least **100 N** force is needed to lift the cube.