9.00 Kg Rock Whose Density Is 4500 Kg/m3 Is Suspended By A String Such That Half Of The Rock's Volume

9.00 Kg Rock Whose Density Is 4500 Kg/m3 Is Suspended By A String Such That Half Of The Rock's Volume

Understanding the physical principles behind the suspension of an object such as a rock involves exploring concepts like density, buoyancy, weight, and tension. In this article, we delve into the details of a 9.00 kg rock with a density of 4500 kg/m³, suspended by a string in such a way that half of its volume is submerged or supported by the string. We will analyze the problem step by step, explaining the relevant physics concepts and calculations involved.

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Introduction to Density and Mass

What Is Density?

Density is a measure of how much mass is contained in a unit volume of a substance. It is expressed mathematically as:

• Density (ρ) = Mass (m) / Volume (V)

In our case, the density of the rock is given as 4500 kg/m³, which indicates that each cubic meter of this rock weighs 4500 kilograms.

Mass and Its Relationship to Density and Volume

The mass of the rock is provided as 9.00 kg. Using the density, we can determine the volume of the rock:

V = m / ρ

Calculating this provides insight into the size of the rock, which is essential for understanding the mechanics of its suspension.

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Calculating the Volume of the Rock

Step-by-Step Volume Calculation

Given:

• Mass, m = 9.00 kg
• Density, ρ = 4500 kg/m³

Applying the formula:

V = m / ρ = 9.00 kg / 4500 kg/m³ = 0.002 m³

This means the total volume of the rock is 0.002 cubic meters, or 2 liters.

Implications of the Volume

Knowing the volume helps us understand how much of the rock is submerged or supported when suspended, especially since the problem states that half of the volume is involved in the suspension dynamics.

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Understanding the Suspension and Buoyancy

Basic Principles of Buoyancy

When an object is submerged in a fluid (or suspended in a medium like air with supporting forces), it experiences a buoyant force equal to the weight of the displaced fluid:

• Buoyant Force (Fb) = ρfluid × g × V_submerged

Where:

• ρ_fluid is the density of the fluid or medium


• g is acceleration due to gravity (~9.81 m/s²)


• V_submerged is the volume of the object submerged

In our case, since the rock is suspended by a string, the key forces are its weight and the tension in the string.

The Role of the String and Half-Volume Suspension

The problem states that the rock is suspended such that half of its volume is involved. This can mean:

• Half of the rock's volume is submerged in a fluid, or
• The tension in the string supports the rock in a way that effectively involves half of the volume in buoyant support.

Assuming the latter, the tension in the string balances the weight minus the buoyant force acting on the submerged part of the rock.

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Analyzing the Forces Acting on the Rock

Forces Involved

The main forces acting on the suspended rock are:

• The gravitational force (its weight): \(W = m \times g\)
• The buoyant force (if submerged): \(F_b\)
• The tension in the string: \(T\)

The equilibrium condition (assuming the rock is stationary) is:

\[ T + F_b = W \]

Calculating the Weight of the Rock

Given:

• m = 9.00 kg
• g ≈ 9.81 m/s²

Weight:

\[ W = 9.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 \approx 88.29 \, \text{N} \]

This is the downward force due to gravity.

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Determining the Buoyant Force and Submerged Volume

Scenario 1: Half of the Volume Is Submerged

If exactly half of the volume of the rock is submerged:

• V_submerged = V / 2 = 0.001 m³

Assuming the medium surrounding the rock is air, and its density is approximately 1.2 kg/m³, the buoyant force is:

\[ Fb = ρ{air} \times g \times V_{submerged} \]

Calculating:

\[ F_b = 1.2 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.001 \, \text{m}^3 \approx 0.0118 \, \text{N} \]

This is negligible compared to the weight (~88.29 N), so in air, buoyancy doesn't significantly affect the tension.

Scenario 2: Submerged in a Fluid with Higher Density

If the rock is submerged in a fluid like water (density ≈ 1000 kg/m³):

\[ F_b = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.001 \, \text{m}^3 = 9.81 \, \text{N} \]

Again, this reduces the effective weight supported by the tension:

\[ T = W - F_b \]

\[ T = 88.29 \, \text{N} - 9.81 \, \text{N} = 78.48 \, \text{N} \]

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Implications and Practical Understanding

Why Does Half the Volume Matter?

The core aspect of the problem is understanding the significance of "half of the volume" in suspension:

• If half of the volume is submerged in a fluid with significant density, it reduces the tension in the string.
• If the object is in air, the buoyant effect is minimal, and the tension is nearly equal to the weight.

Applications in Real-World Situations

Understanding how objects behave when suspended or submerged is crucial in:

• Designing ships and submarines
• Engineering buoyancy aids
• Analyzing forces in physics experiments involving fluids

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Summary of Key Calculations

• Volume of the rock: 0.002 m³
• Weight of the rock: approximately 88.29 N
• Submerged volume (half): 0.001 m³
• Buoyant force in air: negligible (~0.012 N)
• Buoyant force in water: approximately 9.81 N
• Tension in the string (in water): approximately 78.48 N

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Conclusion

In summary, a 9.00 kg rock with a density of 4500 kg/m³ has a total volume of 0.002 m³. When suspended such that half of its volume is involved in the support system, the physics of buoyancy and weight come into play. In air, buoyant forces are negligible, so the tension in the string nearly equals the weight of the rock. However, if the rock is submerged in a fluid like water, buoyant forces significantly reduce the tension required in the supporting string, illustrating the importance of volume, density, and medium in understanding physical systems. This analysis not only helps in solving theoretical physics problems but also has practical implications in engineering and fluid mechanics.