Block A in Figure 13.45 hangs by a cord from spring balance...

Block $A$ in Figure 13.45 hangs by a cord from spring balance $D$ and is submerged in a liquid $C$ contained in beaker $B$ . The mass of the beaker is $1.00 \mathrm{kg} ;$ the mass of the liquid is 1.80 $\mathrm{kg} .$ Balance $D$ reads 3.50 $\mathrm{kg}$ and balance $E$ reads 7.50 $\mathrm{kg} .$ The volume of block $A$ is $3.80 \times 10^{-3} \mathrm{m}^{3}$ . (a) What is the density of the liquid? (b) What will each balance read if block $A$ is pulled up out of the liquid?

Key Concepts

Apparent Weight

The apparent weight of an object is the weight it seems to have when immersed in a fluid, which is less than its actual weight due to the upward buoyant force. Understanding the difference between apparent and true weight is key to interpreting scale readings and solving problems involving submerged objects.

Archimedes' Principle

This principle states that any object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object. It is crucial for understanding how the net force on a submerged object is reduced compared to its weight in air and for calculating forces in systems where objects are submerged.

Density

Density is defined as the mass per unit volume of a substance. It is essential for determining the properties of both the liquid and any submerged objects. In problems like this, calculating the density of the liquid helps in relating the mass of the displaced liquid to the buoyant force exerted on the submerged object.

Force Balance and Equilibrium

Force balance and equilibrium involve ensuring that the sum of forces acting on a system is zero for the system to be static. This concept is used to analyze the readings on different balances when forces such as weight, buoyant force, and tensions in cords interact, ensuring that all forces are properly accounted for in both submerged and non-submerged scenarios.

Transcript

00:03 All right, so the starting observation for this problem is that the scale that the beaker is sitting on is going to read a value that's higher than the mass of the beaker plus the mass of the liquid.

00:20 And the reason why that value is higher is because the liquid is exerting a buoyant force on the block.

00:28 And by newton's second law, then the block has to be exerting an equal and opposite force down on the liquid in the beaker.

00:37 So that adds to the reading on the scale.

00:40 And we can use this fact to deduce what the density of the liquid is.

00:47 So what we want to do is set the force that the bottom scale reads, which is 7 .5 kilograms times the ground.

01:00 Gravitational acceleration g that's going to be equal to the weight of the beaker plus the weight of the liquid plus the buoyant force.

01:16 So if we divide both sides of this equation by g, we can draw all the g's and then on the right here we have f buoyant divided by g.

01:38 By g is also equal to, by archimedes principle, equal to the density of the liquid times the volume of the block.

01:50 And the volume of the block was given to us at 3 .8 times 10 to the negative 3 meters cubed.

01:56 So we solve this first equation for f buoyant over g.

02:07 That's going to be 7 .5 minus the mass of the beaker, which is 1 .5.

02:17 Minus the mass of the liquid, which is 1 .8.

02:23 Then after we have this f buoy over g, we can set that equal to row liquid times v block and solve this for row liquid...

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