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Chapter 14: Problem 33

An iron anchor of density \(7870 \mathrm{~kg} / \mathrm{m}^{3}\) appears \(200 \mathrm{~N}\) lighter in water than in air. (a) What is the volume of the anchor? (b) How much does it weigh in air?

Short Answer

Expert verified
Volume is approximately 0.0204 m鲁; weight in air is approximately 1572.88 N.

Step by step solution

01

Understand the Problem

We have an iron anchor that appears lighter in water due to the buoyant force. We need to find the volume of the anchor and its weight in the air.
02

Recall Archimedes' Principle

According to Archimedes' principle, the buoyant force equals the weight of the displaced fluid. The anchor appears 200 N lighter, so the buoyant force is 200 N.
03

Calculate the Volume of the Anchor

The buoyant force in water is equal to the weight of the water displaced, which is given by the equation \( F_b = \rho_{water} \times V \times g \), where \( F_b = 200 \) N, \( \rho_{water} = 1000 \) kg/m鲁, and \( g = 9.8 \) m/s虏. Solving for \( V \), we get \[ V = \frac{F_b}{\rho_{water} \times g} = \frac{200}{1000 \times 9.8} \approx 0.0204 \text{ m}^3 \].
04

Calculate the Mass of the Anchor

With the volume of the anchor known, use the density of iron to find its mass. The equation \( m = \rho_{iron} \times V \) is used, with \( \rho_{iron} = 7870 \) kg/m鲁 and \( V = 0.0204 \) m鲁. We find \[ m = 7870 \times 0.0204 \approx 160.498 \text{ kg} \].
05

Calculate the Weight of the Anchor in Air

The weight of the anchor in air is the force due to gravity and is given by \( W = m \times g \). Using \( m = 160.498 \) kg and \( g = 9.8 \) m/s虏, we calculate \[ W = 160.498 \times 9.8 \approx 1572.88 \text{ N} \].

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Key Concepts

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

Buoyant Force
When an object is submerged in a fluid, like water, it experiences an upward force called buoyant force.
This force is responsible for making objects appear lighter when they are underwater. Have you ever felt like you weighed less while swimming? That's all because of buoyant force.
According to Archimedes' Principle, this buoyant force is equal to the weight of the fluid displaced by the object.
  • If the object displaces a lot of water, the buoyant force is strong.
  • If it displaces very little, the force is weak.
For instance, in the problem with the anchor, the buoyant force is measured at 200 N. This information tells us how much water weight the anchor is displacing. By knowing the buoyant force, we can derive other important properties like volume.
Density Calculation
Density is a key concept in understanding why objects float or sink. It is defined as the mass per unit volume of an object, usually expressed in kg/m鲁.
In our exercise, the density of iron is given as 7870 kg/m鲁.
This high density suggests that iron is much heavier compared to water and explains why it sinks. To calculate an object's density, you need two things: its mass and its volume.
  • Density formula: \( \rho = \frac{m}{V} \)
  • The higher the density, the less likely an object is to float.
In the example, we initially used the density formula to find the anchor's mass after having determined its volume through the buoyant force. It's fascinating to see how these concepts interconnect!
Weight Measurement
Weight is the measure of the gravitational force acting on an object's mass. It describes how heavy something feels when you lift it.
The weight of an object can be calculated using the formula: \( W = m \times g \), where \( g \) is the acceleration due to gravity, approximately 9.8 m/s虏 on Earth.
When you weigh an object in the air, you are essentially measuring this gravitational force without the interference of buoyancy. For example, in the exercise, after calculating the mass of the anchor, we found the weight in air to be around 1572.88 N.
  • This demonstrates how weight can differ in various environments; it's influenced by factors like buoyant force when submerged.
  • Weight will always feel less in a liquid due to the upward buoyant force acting against gravity.
Understanding these concepts clarifies how weight measurement can vary depending on the situation.

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Most popular questions from this chapter

An iron casting containing a number of cavities weighs 6000 \(\mathrm{N}\) in air and \(4000 \mathrm{~N}\) in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is \(7.87 \mathrm{~g} / \mathrm{cm}^{3}\).

A liquid of density \(900 \mathrm{~kg} / \mathrm{m}^{3}\) flows through a horizontal pipe that has a cross-sectional area of \(1.90 \times 10^{-2} \mathrm{~m}^{2}\) in region \(A\) and a cross-sectional area of \(9.50 \times 10^{-2} \mathrm{~m}^{2}\) in region \(B\). The pressure difference between the two regions is \(7.20 \times 10^{3} \mathrm{~Pa}\). What are (a) the volume flow rate and (b) the mass flow rate?

Suppose that you release a small ball from rest at a depth of \(0.600 \mathrm{~m}\) below the surface in a pool of water. If the density of the ball is \(0.300\) that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)

A simple open U-tube contains mercury. When \(11.2 \mathrm{~cm}\) of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm?

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density \(1.30 \times 10^{3}\) \(\mathrm{kg} / \mathrm{m}^{3}\). The area of each base is \(4.00 \mathrm{~cm}^{2}\), but in one vessel the liquid height is \(0.854\) \(\mathrm{m}\) and in the other it is \(1.560 \mathrm{~m}\). Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.
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