/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A uniform \(\operatorname{rod} A... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform \(\operatorname{rod} A B, 12 \mathrm{~m}\) long weighing \(24 \mathrm{~kg}\), is supported at end \(B\) by a flexible light string and a lead weight (of very small size) of \(12 \mathrm{~kg}\) attached at end \(A\). The rod floats in water with one-half of its length submerged. Find the volume of the rod (in \(\times 10^{-3} \mathrm{~m}^{3}\) ) [Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), density of water \(\left.=1000 \mathrm{~kg} / \mathrm{m}^{3}\right]\)

Short Answer

Expert verified
The volume of the rod is 144.

Step by step solution

01

Determine Buoyant Force

The rod is floating with half of its length submerged in water. According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced by the submerged portion of the rod. Thus, the buoyant force is equal to the weight of the rod plus the additional weight added to it. Since the rod weighs 24 kg and an additional 12 kg weight is attached, the total weight supported by buoyancy is \[ (24 + 12) \times g = 36 \times 10 = 360 \text{ N}. \]
02

Calculate Displaced Volume of Water

The buoyant force is also determined by the weight of the displaced water, which is the volume of the submerged part of the rod times the density of water times gravity. Since half the rod is submerged, we have: \[ \text{Buoyant force} = \text{Volume submerged} \times \text{density of water} \times g. \]Since half of the rod's length is submerged, the submerged volume is half of the total volume. Hence: \[ 360 = \left( \frac{1}{2} V \right) \times 1000 \times 10. \]
03

Solve for the Rod's Total Volume

Solving for the rod's total volume \(V\), we have:\[ 360 = 5000 \times \left( \frac{V}{2} \right), \]\[ 360 = 2500V, \]\[ V = \frac{360}{2500} = 0.144 \text{ m}^3. \]Thus, the volume of the rod is \(0.144 \times 10^{-3} \text{ m}^3 \) when expressed in required units.

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Key Concepts

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

Buoyant Force
The concept of buoyant force is central to understanding why objects float. According to Archimedes' Principle, an object submerged in a fluid experiences an upward force that is equal to the weight of the fluid it displaces. This upward push is what we call the buoyant force.
For instance, in the problem involving the rod, the rod's ability to float in water with half its length submerged is a perfect demonstration of this principle. Here, the buoyant force not only supports the weight of the rod, which is 24 kg but also the 12 kg weight attached to it. This total of 36 kg multiplied by gravity gives us a buoyant force of 360 N.
Understanding buoyancy helps clarify why some objects float effortlessly while others sink. It's all about balancing the weight of the object with the weight of the fluid displaced.
Displaced Volume
Displaced volume refers to the amount of fluid that an object pushes aside when it is submerged in a fluid. In the case of the rod from the exercise, half of its length is submerged in water, meaning it displaces a volume of water equal to half of its total volume.
This submerged portion is crucial as it determines the volume of water displaced and consequently the magnitude of the buoyant force acting on the rod. When the rod displaces water, it creates a force that supports it against gravity, allowing it to float.
  • This concept is especially vital for predicting real-life applications, such as ship design, where the balance of weight and displaced water ensures that the vessel remains afloat.
It's interesting how simply displacing water can generate a powerful force of buoyancy, essential for various engineering feats.
Density of Water
The density of water plays a fundamental role in calculating buoyant force and displaced volume. Density is defined as mass per unit volume and for water, it is typically \(1000\, ext{kg/m}^3\). This property allows us to compute the exact amount of buoyant force an object can experience when in water.
In the exercise, knowing the density of water is key to finding out how much water the submerged half of the rod displaces. Coupling this with gravity gives us the means to directly determine the buoyant force using the formula:
  • Buoyant force = Displaced Volume × Density of Water × Gravity.
By understanding the density of fluid involved, we can accurately predict and control the buoyancy and stability of floating structures, whether they're tiny rodlike objects or enormous ships.
It's evident that the role of water's density is inseparable from any buoyancy calculations in fluid dynamics.

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Most popular questions from this chapter

Equal masses of three liquids are kept in three identical cylindrical vessels \(A, B\) and \(C\). Their densities are \(\rho_{A}, \rho_{s}\) and \(\rho_{c}\) with \(\rho_{A}<\rho_{s}<\rho_{c^{\prime}}\) The force on the base will be (1) maximum in vessel \(A\) (2) maximum in vessel \(B\) (3) maximum in vessel \(C\) (4) same in all the three vessels

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If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid for calculation of pressure, effective \(g\) is used. A closed box with horizontal base \(6 \mathrm{~m}\) by \(6 \mathrm{~m}\) and a height \(2 \mathrm{~m}\) is half filled with liquid. It is given a constant horizontal acceleration \(g / 2\) and vertical downward acceleration \(g / 2\). What is the value of vertical acceleration of box for given horizontal acceleration \((g / 2)\), so that no part of the bottom of the box is exposed? (1) \(g / 2\) upwards (2) \(\mathrm{g} / 4\) downwards (3) \(g / 4\) upwards (4) not possible

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