# Buoyancy question #2

• October 16th, 2006, 10:08 AM
Liam90
Buoyancy question #2
The apparent weight of a wooden cube floating on water = 0.
Hence, the weight of a cup of water shown on the measuring machine will not change if a wooden cube is added into the water. (true or false?)

What about the apparent weight of a layer of oil floating on water.
Will the weight of a cup of water shown on the measuring machine change if a layer of oil is added on top of it? (if yes, what happened to the buoyancy?)
• October 16th, 2006, 01:10 PM
billiards
Why is the apparent weight for this wooden block zero? It doesn't sink because the buoyancy force is equal and opposite to the weight. The mass per unit volume (or if you like the density) of the wooden block, is less than the water surrounding it, this is the origin of the buoyancy that causes it to float. Nevertheless the wooden block still has weight, adding the block will contribute mass to the system and this translates to an overall increase in weight (as detected by your scales).

Remember the buoyancy acts on the wooden block, not the whole cup!!! As such the buoyancy effect will be internal within the system and will not be detected by the scales which measures the weight of the system as a whole. This applies with the oil in the same way.

The buoyancy is there, but the scales know nothing about it. If you add mass to something without taking any away, the weight will inevitably increase.
• October 17th, 2006, 02:56 AM
lince!
hey liam,
when the wooden block is at equilibrium, the weight does not disspear...it is the NET FORCE which is ZERO.
your question has something to do with dynamics and newton's law. i think you are not very clear about them.
• October 17th, 2006, 11:30 AM
Liam90
Wait, before we go into further discussions, lets get a few things confirmed.

Weighing machines or weighing scales only weigh the apparent weight and not the actual weight right?

Hence, an object with the weight of 20N and an upward force of 15 N will only weigh 5N on the weighing machine because the magnitud of the reaction force (apparent weight) is only 5 N right?

The wooden block has 2 forces acting upon it, weight and buoyancy. The wooden block is floating, weight = buoyancy, thus there is no reaction force and no apparent weight.

What i don't understand is, how can the readings of the weighing scale increase the wooden block has no apparent weight...?
• October 17th, 2006, 03:32 PM
billiards
The apparent weight is the weight because the scales aren't accelerating.
• October 17th, 2006, 03:46 PM
Quote:

Originally Posted by Liam90
Wait, before we go into further discussions, lets get a few things confirmed.

Weighing machines or weighing scales only weigh the apparent weight and not the actual weight right?

Hence, an object with the weight of 20N and an upward force of 15 N will only weigh 5N on the weighing machine because the magnitud of the reaction force (apparent weight) is only 5 N right?

The wooden block has 2 forces acting upon it, weight and buoyancy. The wooden block is floating, weight = buoyancy, thus there is no reaction force and no apparent weight.

What i don't understand is, how can the readings of the weighing scale increase the wooden block has no apparent weight...?

If your 20N is producing an upward force of 15N, then what is it pushing against? - answer, what ever is underneath it. Go and have a look at newtons laws of motion which will show that motion is a result of an imbalance in forces. You are weighing wood + water the result will be the sum.
• October 18th, 2006, 12:53 AM
lince!
eh.........are you talking about pressure and normal force?
• October 18th, 2006, 08:10 AM
sramanujam
The weight of cup of water on measuring instrument will not change as long as buyoncy force is balanced by weight of wooden cube. However, when net force is non-zero, measuring instrument will show you the net force that forces the wooden cube to sink. Same is true for the layer of oil in water.