Dynamics and displacement problem

[aiglosicicle]

hello i have trouble understanding this problem... hope u guys can help. here's the question:

'when a beaker of water rests of a balance, the weight indicated in X. a solid object of weight Y in air displaces water of weight Z, when it is floating. what will be the final reading of the balance?'

according to the answer the reading should be X+Y+Z

I understand that X should be included (duh) but I am confused about Y and Z. Why is the weight of the water displaced included? The water that is displaced comes from the water that is already in the beaker so shouldn't its weight be already included in the X? Also i understand that for an object to float upthrust must be equals to the weight of the object, if they are equal shouldn't the two forces cancel each other out and hence, there is no Y?

 

[rcgldr]

Assume you have 20lbs of beaker and water, and you add 1 lb of ice (air weight) to the beaker.

Water is about 800 times as dense as air (dry air at sea level); ice is about 736 times as dense, so the buoyancy of the air on either is mostly insignificant in this case (less than 1%).

So the bottom line is you have very close to 1lb mass of ice (within 1%), added to the 20lbs of beaker and water. Gravity is pulling downwards on 21lbs of mass, so the scale should read 21lbs, the total of the beaker, water and ice (X+Y).

Assuming the ice is 8% less dense than water (this depends on the temperature of both), then the ice displaces about .92 lbs of water, but the total weight isn't going to be 21.92 lbs.

 

[kyle8921]

Hello,

I can understand your confusion regarding this problem. Let me try to explain it in a simpler way.

When a beaker of water is placed on a balance, the weight indicated is X. This weight includes the weight of the beaker, the water inside the beaker, and the air above the water. Now, when you place a solid object (weight Y) in the water, it displaces some of the water (weight Z). This means that the water level in the beaker will rise, and some of the air above the water will be pushed out.

Now, let's look at the forces acting on the beaker and its contents. The downward force is the weight of the beaker, water, and the solid object (X+Y). The upward force is the buoyant force, which is equal to the weight of the water displaced by the solid object (Z).

Since the object is floating, it means that the buoyant force is equal to the weight of the object (Y). This is because the object is displacing an amount of water that is equal to its own weight.

So, when we add all the forces together, we get X+Y+Z as the final reading on the balance. This is because the weight of the beaker and water (X) is still acting downward, the weight of the floating object (Y) is acting upward, and the weight of the displaced water (Z) is also acting upward.

I hope this explanation helps you understand the problem better. If you have any further questions, please let me know.