91Ó°ÊÓ

A partly filled beaker of water sits on a scale, and its weight is \(2.30 \mathrm{~N}\). When a piece of metal suspended from a thread is totally immersed in the beaker (but not touching bottom), the scale reads \(2.75 \mathrm{~N}\). What is the volume of the metal? The water exerts an upward buoyant force on the metal. According to Newton's Third Law of action and reaction, the metal exerts an equal downward force on the water. It is this force that increases the scale reading from \(2.30 \mathrm{~N}\) to \(2.75 \mathrm{~N}\). Hence the buoyant force is \(2.75-2.30=0.45 \mathrm{~N}\). Then, because $$F_{B}=\text { weight of displaced water }=\rho_{w} g V=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(V)$$ we have the volume of the displaced water, and of the piece of metal, namely, $$ V=\frac{0.45 \mathrm{~N}}{9810 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}^{2}}=46 \times 10^{-6} \mathrm{~m}^{3}=46 \mathrm{~cm}^{3} $$

Step by step solution

01

Determine Buoyant Force

To find the volume of the metal, first calculate the buoyant force exerted by the water. This force also causes the increase in the scale reading from 2.30 N to 2.75 N. The buoyant force is the difference in scale measurement: \[ F_B = 2.75 ext{ N} - 2.30 ext{ N} = 0.45 ext{ N} \]

02

Apply the Buoyant Force Formula

The buoyant force can also be expressed in terms of the weight of the water displaced by the submerged metal, according to the formula: \[ F_B = \rho_w g V \]where \( \rho_w = 1000 \text{ kg/m}^3 \) (density of water), \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity), and \( V \) is the volume of the water displaced, which equals the volume of the metal.

03

Solve for Volume

Substitute the known values into the buoyant force formula to solve for \( V \):\[ 0.45 ext{ N} = (1000 ext{ kg/m}^3)(9.81 ext{ m/s}^2)(V) \]\[ V = \frac{0.45 ext{ N}}{9810 ext{ N/m}^3} \]\[ V = 46 \times 10^{-6} ext{ m}^3 \] This volume can also be converted to \( 46 \text{ cm}^3 \), since \( 1 ext{ m}^3 = 10^6 ext{ cm}^3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Displacement

When an object is fully submerged in a fluid, it pushes aside, or displaces, a volume of that fluid equal to the volume of the object itself. This is the principle of volume displacement. Understanding this concept helps explain why an object seems to lose weight when submerged.
In the original exercise, a piece of metal is immersed in water, causing the water to exert an upward force that can be measured. This force is directly related to the volume of water displaced by the metal.

• The displaced water takes up space, and the amount of volume it occupies equals the volume of the object.
• This displaced volume can be determined from the increase in the scale's weight reading.

Recognizing how volume displacement works is critical to solve the exercise, as it allows us to relate physical object volume to measurable forces.

Buoyant Force Calculation

Buoyant force is the net upward force experienced by an object submerged in a fluid. This force is a result of pressures exerted by the fluid and increases with the density of the fluid and the volume of the object submerged. When you subtract the apparent weight of the object when submerged from its weight in air, you can find the buoyant force.
In the exercise, calculating the change in scale reading from 2.30 N to 2.75 N gives us the buoyant force:

• The scale reads higher by 0.45 N when the metal is dipped into the water.
• This increase matches the buoyant force acting upward on the metal.

Formally, the buoyant force \( F_B \) is calculated using \( F_B = \rho_w g V \), where \( \rho_w \) is the density of water, \( g \) is gravitational acceleration, and \( V \) is the volume of displaced water. Plugging these values in, you solve for volume, revealing the size of the metal piece.

Newton's Third Law

Newton's Third Law states that for every action, there is an equal and opposite reaction. In the context of buoyancy, this principle explains the interaction between the weight of the displaced fluid and the object immersed in it. The weight of the fluid displaced by the object generates a buoyant force that pushes the object upwards.
In our beaker scenario:

• The weight of the water displaced by the metal (acting downward) is counterbalanced by the buoyant force (acting upward).
• The metal piece exerts a downward force on the water, and by Newton's Third Law, the water exerts an equal and opposite force on the metal.

This balancing of forces highlights how the changes in the weighing scale reflect the net force generated, ensuring the metal remains suspended but does not sink. Understanding this law is essential for grasping the underlying dynamics of buoyancy.