F = ma 2012 #20 (Apparent weight when dunking wood in water)

[SignaturePF]

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?

 

[Simon Bridge]

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

Every action has an equal and opposite reaction.
Draw the free-body diagram.

 

[SignaturePF]

Ok so we have F_b up and normal force of the scale up. So the scale will read 55kg?

 

[Simon Bridge]

The wood experiences the buoyancy force up and gravity down... is all the weight of the wood supported by the buoyancy force?

The water experiences the buoyancy force and gravity down, and the scales up to match.

You've actually answered these questions: I'm going for understanding.
What if the wood were floating in the water instead of being suspended in it?
What if it were pushed under the water so it was completely submerged?

 

[SignaturePF]

If the wood were completely floating in the water; then F_b = 0, if a fraction, then buoyancy force is βmg, where B is the fraction under water.
If the wood were pushed under the water; Fb > Fg, and the difference is the force of the push.

 

[Simon Bridge]

Try again.

When an object is floating, it is being supported somehow... otherwise we say it is "sinking". An object floats when this force is equal and opposite the object's weight. It is not going to be zero.

The supporting force depends on the relative density of the object and fluid.
It may float with part sticking out of the water, or, if the relative density is 1, at any depth.

When it is floating, it displaces it's mass of the fluid.
When submerged, it displaces it's volume of fluid.