Question

A steel cube 0.39 m on each side is suspended from a scale and immersed in water. What will the scale read? In N

Answer 1

Side of the cube a = 0.39 m

Volume of the cube V = a ³

                                 = 59.319 x10 ^-3 m ³

Density of the steel D = 7800 kg /m ³

Mass M = DV= 462.68 kg

Weight of the cube W = Mg

                                 = 462.68 x9.8

                                 = 4534.34 N

Buoyancy force F = V d g

Where d = density of water = 1000 kg / m³

So, F = 59.319 x10 ^-3 x1000 x 9.8 = 581.32 N

Net force on the cube F ' = W - F

                                      = 4534.34 N - 581.32 N

                                       = 3953.02 N

                                       ~ 3953 N

Note : if the scale is calibrated in kg then the sclae reading is = 3953 N / 9.8 m/s ²

                                                                                           = 403.36 kg