### A101 6P P03 Hang Float and Sink

```A101 Science
Problem 03: Hang, Float and Sink
6th Presentation
Understanding the problem
•
Before water touched the bottom of
the hanging object, the weighing
scale reading for both setups were
the same when the same amount of
?
?
Set-up 1
Set-up 2
•
The reading was equal to the sum of the mass of the
container and the mass of water added.
•
If we continue adding water, will the weighing scale
readings for the two setups still remain equal?
Observations
h
?
Set-up 1
?
Set-up 2
Even though we added
the same amount of water
to each container, the
weighing scale reading for
the container with the
partially submerged object
was greater.
h
?
?
Set-up 1
Set-up 2
However, when the water
levels in both containers
were the same, experiment
showed that the weighing
scale readings were the
same even though there
was more water in set-up 1.
Deduction
• Based on the experiment, we can make a
preliminary deduction that with the same height
of the water level, the weighing scale reading
would be the same, regardless of the amount of
water added or the presence of any partially
submerged object.
• The deduction is still valid, even if we change
the mass and/or the shape of the hanging
object.
A possible explanation
• A container has no way of knowing what is immersed in
the water; it can only feel a downward ‘push’ from the
water which is dependent on the height of the water level.
• Once the height of the water levels are the same in both
cases, both containers will experience the same amount
of 'push' from the water, so both weighing scales will end
up showing the same reading.
• It is possible to have other valid explanations for the
deduction.
Accounting for the weighing scale
• Volume of submerged portion of object = 10 cm x 10 cm x 2 cm = 200 cm3
• Equivalent volume of water = 200 cm3
• Mass of equivalent volume of water = Volume x density
= 200 cm3 x 1 g/cm3
= 200 g
water of
200 cm3
Volume
= 200 cm3
Same weighing
Reading on the weighing scale (g)
Graph of reading (g) vs water added (g)
Once the object is fully
immersed, the weight
of added water gets
duly reflected on the
weighing scale.
?
3000
Contribution from
hanging object:
Volume occupied
by object equivalent
to 500 g of water
Deflected path
due to the
presence of
object
2000
?
Actual Contribution
from water to weighing
scale reading = 1500 g
Apparent mass
2000 g
?
1000
1500 g of water added
Initial
400 g 0
1000
Water level touching the
bottom surface of the object
2000
Water fully immerses
the object
3000
Hanging
No Object
Mass of water
Case of floating
• In some cases, the equivalent mass of water
due to the volume submerged by some partially
submerged objects is equal to the mass of the
object itself.
• When this happens, the object floats.
• Hence, for floating objects like wood, there is a
maximum volume which is submerged in water.
• This usually occurs when the density of the
object is lower than that of water.
Case of floating
No tension
Cord
slackens;
in the
object
cord floats
now
Object
Maximum volume
object can be
immersed in water
Reading on the weighing scale (g)
?
3000
Response if the object
does not float
Object floats when more water is added.
Object begins
to float.
in weight the object can
make is 200 g
2000
?
?
1000
Initial
400 g 0
600 g of
1000
Water level touching the
bottom surface of the object
Hanging
Floating
No Object
2000
3000
Mass of water
Going further
•
When the string holding the suspended object is cut
and the object sinks to the base of the container, the
reading on the weighing scale is equal to the sum of
mass of the object, the water added and the container,
because the full mass of the object acts on the
weighing scale.
•
For example, when a 500 g object is hung by a string
and fully submerged in a 400 g container containing
400 g of water, the weighing scale reading is 1100 g.
When the string is cut and the object sinks to the base
of the container, the weighing scale reading is 1300 g.
Learning Points
• Based on our experiment in the video, the immersion of the object
into the water causes an increase in the weighing scale reading.
• Comparing two setups, one with and the other without a hanging
object, we found that as long as the height of the water levels are
equal the weighing scale readings will be equal.
• The volume of the submerged portion of the object causes an
increase in weighing scale reading. This increase is equal to the
mass of the same volume of the liquid.
• To explain the phenomenon observed in the video, we made use of
various skills such as making observations and deductions, testing
and verifying results and drawing conclusions.
Discussion
Two cylindrical containers with different radius
are being filled with water as shown below. The
reading on Weighing Scale C is ______ the
reading on Weighing Scale D.
a. more than
b. less than
c. equal to
```