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Chapter 11: Problem 40

The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3},\) and the density of sea water is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). A swimming polar bear climbs onto a piece of floating ice that has a volume of \(5.2 \mathrm{~m}^{3}\). What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

Short Answer

Expert verified
The ice can support a polar bear weighing up to 554.68 kg.

Step by step solution

01

Understand the Problem

We need to calculate the maximum weight a piece of ice can support while still floating. The key is knowing how buoyancy works: the weight of the displaced water must equal the total weight of the ice and the bear.
02

Calculate the Weight of the Ice

The mass of the ice is the product of its volume and density: \( \text{mass of ice} = \text{volume} \times \text{density} = 5.2 \mathrm{~m}^3 \times 917 \mathrm{~kg/m}^3 \). Thus, the mass of the ice is \(4768.4 \mathrm{~kg}\). The weight of the ice is \( \text{mass} \times g = 4768.4 \mathrm{~kg} \times 9.81 \mathrm{~m/s}^2 \), which is approximately \(46817.004 \mathrm{~N}\).
03

Calculate the Displacement of Water

The volume of water displaced is equal to the volume of the ice, which is \(5.2 \mathrm{~m}^3\). The mass of the displaced water is \(5.2 \mathrm{~m}^3 \times 1025 \mathrm{~kg/m}^3 = 5330 \mathrm{~kg}\). Hence the weight of the displaced water is \(5330 \mathrm{~kg} \times 9.81 \mathrm{~m/s}^2 = 52257.3 \mathrm{~N}\).
04

Determine the Weight of the Bear

Since for the ice to just float without sinking, the total weight (ice plus bear) should equal the weight of the displaced water. So, \(\text{weight of bear} = 52257.3 \mathrm{~N} - 46817.004 \mathrm{~N} = 5440.296 \mathrm{~N}\).
05

Conclusion

The heaviest bear that the ice can support without sinking is 5440.296 N. This means the bear can weigh up to \(\frac{5440.296 \mathrm{~N}}{9.81 \mathrm{~m/s}^2} \approx 554.68 \mathrm{~kg}\).

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Key Concepts

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

Density: The Buoyancy Game Changer
Density is a fundamental concept in buoyancy. It's the mass per unit volume of a substance. In our scenario, we have ice and seawater, each with their own densities. Density determines how these materials interact in water.
For ice, the density is lower than seawater's at 917 kg/m鲁 compared to 1025 kg/m鲁. This difference is crucial. The lower density of ice allows it to float.
It means that per volume unit, ice is lighter than seawater.
Importantly, if a substance has a density lower than the fluid it's placed in, it will float. That鈥檚 why icebergs float! Understanding this concept helps explain why a polar bear can climb onto floating ice without it sinking immediately鈥攂ecause the ice is less dense than the water around it.
Weight: Balancing Forces in Water
Weight is the force exerted by gravity on an object and is calculated by multiplying the mass of the object by the gravitational acceleration (9.81 m/s虏 on Earth). In buoyancy problems, the weight of both the floating object and fluid displaced plays a critical role.
Using our example, the ice weighing 46817.004 N already pushes against the force exerted by the water below.
  • The weight of the ice is calculated using its mass, derived from its volume and density.
  • Adding the bear's weight means increasing the load on the ice.
  • This total mustn't exceed the upward buoyant force for the ice and bear to stay afloat.
So, the heaviest bear that the ice can support safely is determined by finding the balance between these forces.
Displacement: Supporting Weight in Water
Displacement is key to understanding buoyancy. It refers to the amount of water an object pushes aside when it is submerged or partially submerged in a fluid. When the ice floats, its displacement creates an upward force.
This force, known as the buoyant force, must equal the weight of the water displaced by the submerged part of the ice.
  • We determine the displaced water's weight by multiplying its volume by its density and gravitational acceleration.
  • For the bear to stay on the ice without sinking, the ice's displacement must equal or exceed the combined weight of the bear and ice.
Therefore, the water's displacement can support up to 554.68 kg of bear weight, ensuring the ice and bear do not submerge completely.
Floating: How Ice Bears the Load
Floating is the concept of staying on the surface of a liquid without sinking. In essence, an object floats when its upward buoyant force equals its weight. This principle explains why our ice and bear scenario works. The density differential allows the ice to float when unloaded.
As the bear climbs onto the ice, its weight adds to the system, increasing downward forces.
  • Buoyant forces react by pushing upward due to water displacement.
  • The ice continues to float, assuming the total force (weight of ice plus bear) is equal to or less than the buoyant force.
Thus, for the heaviest bear of about 554.68 kg, the ice can still float because the buoyant force matches the combined weight, maintaining the delicate balance that keeps them from sinking entirely.

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Most popular questions from this chapter

A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. For simplicity, consider the weight to be only that of the hot air within the balloon, thus ignoring the balloon fabric and the basket. (a) How is the weight of the hot air determined from a knowledge of its density \(\rho\) hot air and the volume \(V\) of the balloon? (b) For a given volume \(V\) of the balloon, does the buoyant force depend on the density of the hot air inside the balloon, the density of the cool air outside the balloon, or both? Provide a reason for your answer. (c) Draw a free-body diagram for the balloon, showing the forces that act on it. How is the upward acceleration of the balloon related to these forces and to its mass?

A 58 -kg skier is going down a slope oriented \(35^{\circ}\) above the horizontal. The area of each ski in contact with the snow is \(0.13 \mathrm{~m}^{2}\). Determine the pressure that each ski exerts on the snow.

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. It is contained inside a large "medicine dropper," into which the liquid is drawn using a squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). (a) Compared to the weight \(W\) of the tube, how much buoyant force is needed to make the tube float in either battery acid or antifreeze? (b) Is a greater volume of battery acid or a greater volume of antifreeze displaced by the hydrometer to provide the necessary buoyant force? (c) Which mark is farther up from the bottom of the tube? Justify your answers.

A Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=0.0700 \mathrm{~m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{~kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{~m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120 \mathrm{~Pa}\). Find (a) the speed \(v_{2}\) of the gas in the larger original pipe and (b) the volume flow rate \(Q\) of the gas.

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance \(h\), as the drawing shows. It can then be squirted over the meat. (a) Is the absolute pressure in the bulb in the drawing greater than or less than atmospheric pressure? (b) In a second trial, the distance \(h\) is somewhat less than it is in the drawing. Is the absolute pressure in the bulb in the second trial greater or smaller than in the case shown in the drawing? Explain your answers. Using \(1.013 \times 10^{5} \mathrm{~Pa}\) for the atmospheric pressure and \(1200 \mathrm{~kg} / \mathrm{m}^{3}\) for the density of the sauce, find the absolute pressure in the bulb when the distance \(h\) is (a) \(0.15\) \(\mathrm{m}\) and (b) \(0.10 \mathrm{~m}\). Verify that your answers are consistent with your answers to the Concept Questions.
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