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)` .

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial