Question

1. Water is in the big beaker in the figure on the left. Scale 1 reads 58 newtons, scale 2 reads 908 newtons, and scale 3 reads 0 newtons. The hanging block has a density of 10

Answer 1

Force of gravity on the hanging block = 58 newtons, so its mass = 58/G kg (I use G to represent the force of gravity)

= 58/ 9.81 = 5.9123 kg, or 5912 grams

Its density = 10000 kg/m^3, or (10000 x 1000) / 10^6 g /cm^3 = 10 g /cm^3

Now we can find its volume, V = 5912 / 10 = 591.2cm^3

And now we can apply Archimedes' Principle when the block is lowered into the water :

Upthrust = weight of water displaced

Weight of 591.2 cm^3 water = 591.2g, or 0.59 kg

And 0.96 kg represents 0.59 x 9.81 N = 5.7879 N

So with an upthrust of 5.79 newtons, Scale 1 will read (58 - 5.79) = 52.21 N

This upthrust is transferred directly to Scale 2, which will therefore read
(908 + 5.7879) = 913.788 N

If the valve is open, a volume of water equal to the volume of the block is displaced into the container on Scale 3, and we already know that this is equivalent to 5.7879 newtons, so Scale 3 reads 5.7879.

We have seen that the force registered by Scale 2 increases by the amount of upthrust on the block, that is, 5.7879 N. But this is exactly balanced by the water which flows out into the container alongside. So the reading on Scale 2 remains unchanged at 908 N.