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## Homework Statement

20. A container of water is sitting on a scale. Originally, the scale reads M1 = 45 kg. A block of wood is suspended

from a second scale; originally the scale read M2 = 12 kg. The density of wood is 0.60 g/cm^3; the density of the

water is 1.00 g/cm3

. The block of wood is lowered into the water until half of the block is beneath the surface.

What is the resulting reading on the scales?

45.0 kg ? kg

? kg

12.0 kg

(A) M1 = 45 kg and M2 = 2 kg.

(B) M1 = 45 kg and M2 = 6 kg.

(C) M1 = 45 kg and M2 = 10 kg.

(D) M1 = 55 kg and M2 = 6 kg.

(E) M1 = 55 kg and M2 = 2 kg CORRECT

## Homework Equations

F_apparent = F_g - F_b

F_g = mg

F_b = m_fg

F_b = ρ_fV_fg

F_b = ρVg

## The Attempt at a Solution

I've heard that one easy approach is to consider that the initial net external forces is the same as the final net external forces. So the only answer that makes sense is E -- which sums to 57. Why is this approach valid?

But my approach was as follows:

Wood:

F_app = 12g - (1000 kg / m^3)(6 kg)(1 m^3 / 600 kg)g

F_app = 2 g

m_app = 2kg

Water:

In this case, the apparent weight should be:

F_app = 45g + that mess up there with the densities

F-app = 55 g

m_app = 55 kg

But why should the buoyancy force be added to the weight?