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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

You are doing experiments from a research ship in the Atlantic Ocean. On a day when the atmospheric pressure at the surface of the water is \(1.03 \times 10^{5} \mathrm{~Pa}\), at what depth below the surface of the water is the absolute pressure (a) twice the pressure at the surface and (b) four times the pressure at the surface?

Scientists have found evidence that Mars may once have had an ocean \(0.500 \mathrm{~km}\) deep. The acceleration due to gravity on Mars is \(3.71 \mathrm{~m} / \mathrm{s}^{2}\). (a) What would be the gauge pressure at the bottom of such an ocean, assuming it was freshwater? (b) To what depth would you need to go in the earth's ocean to experience the same gauge pressure?

A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is \(1.20 \times 10^{4} \mathrm{~Pa}\), while in the region of blockage it is \(1.15 \times 10^{4} \mathrm{~Pa}\). Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at \(30.0 \mathrm{~cm} / \mathrm{s}\), and the specific gravity of this patient's blood is 1.06 . What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?

On another planet that you are exploring, a large tank is open to the atmosphere and contains ethanol. A horizontal pipe of cross sectional area \(9.0 \times 10^{-4} \mathrm{~m}^{2}\) has one end inserted into the tank just above the bottom of the tank. The other end of the pipe is open to the atmosphere. The viscosity of the ethanol can be neglected. You measure the volume flow rate of the ethanol from the tank as a function of the depth \(h\) of the ethanol in the tank. If you graph the volume flow rate squared as a function of \(h,\) your data lie close to a straight line that has slope \(1.94 \times 10^{-5} \mathrm{~m}^{5} / \mathrm{s}^{2} .\) What is the value of \(g,\) the acceleration of a free-falling object at the surface of the planet?

At a certain point in a horizontal pipeline, the water's speed is \(2.50 \mathrm{~m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4} \mathrm{~Pa}\). Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.
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