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A rock with density \(1200 \mathrm{~kg} / \mathrm{m}^{3}\) is suspended from the lower end of a light string. When the rock is in air, the tension in the string is \(28.0 \mathrm{~N}\). What is the tension in the string when the rock is totally immersed in a liquid with density \(750 \mathrm{~kg} / \mathrm{m}^{3}\) ?

Short Answer

Expert verified
The tension in the string is the difference between the weight of the rock and the buoyant force on it. The exact value needs to be calculated using the given densities of the rock and fluid, the given tension when the rock is in air, and the known acceleration due to gravity.

Step by step solution

01

Calculate the volume of the rock

Use the formula for density, \(\rho = m/v\), where \(m\) is mass and \(v\) is volume, to find the volume of the rock. Rearrange the formula to get \(v = m/\rho\). We can calculate the mass of the rock using the tension in the string when the rock is in air as the weight of the rock equals the tension in the string. So, \(m = T_{air}/g\) where \(T_{air} = 28.0N\) and \(g = 9.81m/s^2\). Substitute this mass in the volume formula.
02

Calculate the buoyant force

The buoyant force is equal to the weight of the liquid displaced by the rock, which can be calculated using the formula \(F_{B} = \rho_{liq} \cdot g \cdot v\) where \(\rho_{liq} = 750 kg/m^3\) is the density of the fluid, \(g = 9.81m/s^2\) is the acceleration due to gravity and \(v\) is the volume of the rock calculated in step 1.
03

Calculate the tension in the string when the rock is immersed

The tension in the string when the rock is immersed is the difference between the weight of the rock and the buoyant force. So, \(T_{immersed} = m \cdot g - F_{B}\). Substitute the values of \(m\), \(g\), and \(F_{B}\) calculated in the previous steps.

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Key Concepts

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

Density and Buoyancy
Understanding the relationship between density and buoyancy is crucial in many areas of physics, particularly when analyzing objects submerged in fluids. Density is defined as the mass (\(m\)) of an object divided by its volume (\(v\)), expressed as \[ \rho = \frac{m}{v} \.\] Buoyancy, on the other hand, is the upward force that a fluid exerts on an object less dense than itself. This force is why objects float or seem lighter when submerged.

Archimedes’ principle states that the buoyant force on an object is equal to the weight of the fluid it displaces. In mathematical terms, if \( F_B \) is the buoyant force, \( \rho_{liq} \) is the density of the liquid, and \( g \) is the acceleration due to gravity, then \[ F_B = \rho_{liq} \cdot g \cdot v \.\] This means that the amount of fluid displaced, and its density, directly affect the buoyant force—and consequently affect the tension in a string holding a submerged object.

To solve problems involving buoyancy, we often use the densities of the object and fluid to calculate the volume of the object submerged. Then, we apply Archimedes' principle to find the buoyant force, which helps us understand the net forces acting on the object. For students looking to master this concept, they need to focus on the clear understanding of how to manipulate the formula for density, as well as comprehending how buoyant force is calculated and influenced by the properties of the fluid and the object.
Tension in a String
When it comes to examining forces involved in systems where objects are suspended or supported by strings, the concept of tension becomes important. Tension in a string is the force that is transmitted through the string when it is pulled tight by forces acting from opposite ends. In physical problems, tension can be thought of as the 'strength' of the string that counteracts the forces pulling it apart.

For anchored objects, such as a rock suspended by a string, the tension in the string is equal to the force of the weight of the object when it's in the air. The equation representing this situation is \[ T = m \cdot g \.\] However, when the object is submerged in a fluid, the tension changes due to the upward buoyant force. The object feels lighter, and thus the tension in the string is reduced. This new tension can be calculated by taking the original tension and subtracting the buoyant force (\( F_B \) ): \[ T_{immersed} = T_{air} - F_B \.\] Students should recognize that the tension in the string is the result of the balance between gravitational forces and buoyant forces, and that understanding the contributing factors will help them calculate the adjusted tension accurately.

Furthermore, recognizing that the tension in the string provides valuable information about the system's dynamics is essential. It can signify whether the system is in equilibrium and can also be an indicator of the forces at play within the system, which includes the gravitational force, buoyant force, and any other forces affecting the object.
Fluid Mechanics
Fluid mechanics is a branch of physics concerned with the behavior of fluids (liquids and gases) and the forces on them. It is essential in a wide range of applications, from engineering to meteorology. In the context of our problem, fluid mechanics helps us to understand the principles that govern the interaction between the rock and the liquid into which it is immersed.

Key concepts within fluid mechanics involve the calculation of pressure, flow velocity, and the characteristics of different fluid states. In the scenario of the submerged rock, we primarily deal with the principles of static fluid mechanics, also known as hydrostatics. Here, we apply Archimedes’ principle to ascertain the buoyant force experienced by the submerged object.

When dealing with exercises in fluid mechanics, it's crucial for students to comprehend the concepts of density and pressure along with the fact that the pressure at a given depth in a fluid is the same in all directions. Such understanding allows for accurate analysis of the forces in play and forms the basis for calculating tensions and other physical quantities.

By thoroughly grasping the foundational principles of fluid mechanics, students can effectively solve problems not only in their textbooks but also in real-world scenarios involving fluid flows and pressures. This includes predicting how objects will behave when submerged in different fluids, which is a direct application of hydrostatics.

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Most popular questions from this chapter

A pressure difference of \(6.00 \times 10^{4} \mathrm{~Pa}\) is required to maintain a volume flow rate of \(0.800 \mathrm{~m}^{3} / \mathrm{s}\) for a viscous fluid flowing through a section of cylindrical pipe that has radius \(0.210 \mathrm{~m}\). What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to \(0.0700 \mathrm{~m} ?\)

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform - so the hole is \(50.0 \mathrm{~cm}\) above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \(p_{\mathrm{g}}\) of the air in the tank. $$\begin{array}{l|lllll}p_{\mathrm{g}}(\mathrm{atm}) & 0.50 & 1.00 & 2.00 & 3.00 & 4.00 \\\\\hline \boldsymbol{R}(\mathrm{m}) & 5.4 & 6.5 & 8.2 & 9.7 & 10.9\end{array}$$ (a) Graph \(R^{2}\) as a function of \(p_{\mathrm{g}}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in \(h\) during the measurements is very small.

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water completely fills the pipe. (a) At one point in the pipe the radius is \(0.150 \mathrm{~m}\). What is the speed of the water at this point if water is flowing into this pipe at a steady rate of \(1.20 \mathrm{~m}^{3} / \mathrm{s} ?\) (b) At a second point in the pipe the water speed is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the radius of the pipe at this point?

A liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed \(v_{0}\) and the radius of the stream of liquid is \(r_{0}\). (a) Find an equation for the speed of the liquid as a function of the distance \(y\) it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of \(y\). (b) If water flows out of a vertical pipe at a speed of \(1.20 \mathrm{~m} / \mathrm{s},\) how far below the outlet will the radius be one-half the original radius of the stream?

A closed and elevated vertical cylindrical tank with diameter \(2.00 \mathrm{~m}\) contains water to a depth of \(0.800 \mathrm{~m}\). A worker accidently pokes a circular hole with diameter \(0.0200 \mathrm{~m}\) in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of \(5.00 \times 10^{3} \mathrm{~Pa}\) at the surface of the water. Ignore any effects of viscosity. (a) Just after the hole is made, what is the speed of the water as it emerges from the hole? What is the ratio of this speed to the efflux speed if the top of the tank is open to the air? (b) How much time does it take for all the water to drain from the tank? What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

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