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Chapter 12: Problem 37

A large plastic cylinder with mass \(30.0 \mathrm{~kg}\) and density \(370 \mathrm{~kg} / \mathrm{m}^{3}\) is in the water of a lake. A light vertical cable runs between the bottom of the cylinder and the bottom of the lake and holds the cylinder so that \(30.0 \%\) of its volume is above the surface of the water. What is the tension in the cable?

Short Answer

Expert verified
To find the tension in the cable, firstly calculate the buoyant force by finding the weight of the water displaced by the 70% of the cylinder that is submerged. Then subtract this force from the weight of the cylinder to find the tension in the cable.

Step by step solution

01

Calculate the buoyant force

The buoyant force can be found using Archimedes' Principal, which says that the buoyant force on an object is equal to the weight of the fluid displaced by the object. The volume of the cylinder is defined as the mass divided by the density. First, calculate the volume of the whole cylinder using \(V = \frac{m}{\rho}\) where \(m = 30.0 \, \mathrm{kg}\) is the mass of the cylinder and \(\rho = 370 \, \mathrm{kg/m}^3\) is the density of the cylinder. This will give you the volume of the whole cylinder.
02

Calculate the volume under water

We know that 70% of the cylinder is underwater, since 30% of the volume is above the surface. Calculate 70% of the total volume. This is the volume of water displaced, which is needed to calculate the buoyant force.
03

Calculate the weight of the water displaced

Now calculate the weight of the displaced water. The weight of the displaced water is the mass of the displaced water times gravity (g). The mass can be found by multiplying the volume of the displaced water by the density of the water. The density of water is \(1000 \, \mathrm{kg/m}^3\) and \(g = 9.8 \, \mathrm{m/s}^2\).
04

Calculate the weight of the cylinder

The weight of the cylinder is the mass times the gravitational acceleration, \(m \cdot g\). Use the given mass of \(30.0 \, \mathrm{kg}\) and \(g = 9.8 \, \mathrm{m/s}^2\).
05

Find the tension in the cable

Now we can find the tension in the cable. It needs to balance the forces so that the net force on the cylinder is zero. We know that the total force is zero because the cylinder is not moving, therefore the upthrust (buoyant force) plus the tension in the cable must equal the weight of the cylinder. So, the tension (T) can be found using \( T = \mathrm{Weight \, of \, the \, cylinder} - \mathrm{buoyant \, force}\).

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Key Concepts

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

Archimedes' Principle
The Archimedes' Principle is a fundamental concept in physics that explains why objects float or sink in fluids. It states that the buoyant force exerted on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. In other words, an object will experience an upward force when placed in a fluid, and this force is directly proportional to the volume of the fluid displaced by the object.

Application to the Exercise

When applied to our exercise of the plastic cylinder in water, Archimedes' Principle is used to calculate the buoyant force that acts opposite to gravity. We first determine the volume of the cylinder, and then calculate the volume of water displaced by the submerged part of the cylinder. Finally, we find the weight of the displaced water, which tells us the size of the buoyant force. Remember, the buoyant force is what keeps the cylinder afloat and counteracts the weight of the object.
Density and Buoyancy
Density is mass per unit volume and is a measure of how much 'stuff' is packed into a given space. Buoyancy, on the other hand, is the ability of an object to float in a fluid. It is closely related to density because whether an object floats or sinks depends on its density relative to the density of the fluid. An object denser than the fluid will sink, while an object that is less dense will float.

Role in Determining Buoyancy

The cylinder in our example has a known density and mass, allowing us to calculate its volume. Since the density of the cylinder is less than the density of water, the cylinder will float. However, due to the tension in the cable, only a portion of the cylinder's volume is submerged. By calculating the volume of the submerged part, which correlates to the density of the water, we can deduce the cylinder's buoyant force. Understanding the relationship between density and buoyancy is crucial in predicting the equilibrium of submerged objects.
Tension in a Cable
Tension refers to the force transmitted through a string, rope, cable, or any other form of flexible connector when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

Calculating Tension

In our cylinder example, the cable connects the cylinder to the bottom of the lake. The tension in the cable must balance out the net upward buoyant force and the gravitational force (weight) acting on the cylinder to keep it stationary. Mathematically, the tension is found by subtracting the buoyant force from the weight of the cylinder. An essential aspect to remember is that tension is a pulling force, and as such, it can only pull on objects; it cannot push. This concept aids in visualizing and solving problems involving suspended objects, like our partially submerged cylinder.

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Most popular questions from this chapter

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?

A barge is in a rectangular lock on a freshwater river. The lock is \(60.0 \mathrm{~m}\) long and \(20.0 \mathrm{~m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{~N}\) load of scrap metal is put onto the barge. The metal has density \(7200 \mathrm{~kg} / \mathrm{m}^{3}\). (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\). (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}}\) ) and does not mix with it. If water is poured on the surface of that liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{W}}\). (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and \(L=10.0 \mathrm{~cm}\).

On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a 17.7 -m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of \(1600 \mathrm{~kg} / \mathrm{m}^{3}\). If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

You purchase a rectangular piece of metal that has dimensions \(5.0 \times 15.0 \times 30.0 \mathrm{~mm}\) and mass \(0.0158 \mathrm{~kg} .\) The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated?
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