/*! 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 42 (III) A 3.25-kg piece of wood \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(III) A 3.25-kg piece of wood \((\mathrm{SG}=0.50)\) floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink?

Short Answer

Expert verified
A minimum of 3.25 kg of lead is required.

Step by step solution

01

Understand Specific Gravity

The specific gravity (SG) of wood is 0.50. Specific gravity is the ratio of the density of the substance to the density of water. In this case, the density of the wood is half that of water, meaning it will float with half of its volume submerged.
02

Calculate Buoyant Force on Wood

The buoyant force on a floating object is equal to the weight of the fluid displaced. Since the wood floats, the buoyant force is equal to the weight of the wood. Calculate the weight of the wood using the equation: \( \text{Weight of wood} = m_{wood} \times g = 3.25 \times 9.8 = 31.85 \, \text{N} \).
03

Determine Total Volume of Wood

Using the specific gravity, calculate the total volume of the wood. \( \text{Density of water} = 1000 \, \text{kg/m}^3 \), so \( \text{Density of wood} = 0.50 \times 1000 = 500 \, \text{kg/m}^3 \). Therefore, \( V_{wood} = \frac{3.25}{500} = 0.0065 \, \text{m}^3 \).
04

Determine the Volume of Submerged Wood

Since only half the volume is submerged (because SG is 0.50), the submerged volume is: \( V_{submerged} = 0.5 \times 0.0065 = 0.00325 \, \text{m}^3 \).
05

Calculate Buoyant Force Needed to Just Sink the Wood

To sink the wood, the added lead must increase the total submerged volume, so the new buoyant force equates the total weight (wood + lead). Since all volume must submerge, \( V_{total} = V_{wood} \), which means the volume now equals \( 0.0065 \, \text{m}^3 \).
06

Calculate Mass of Lead Required

The volume of the displaced water when the wood (and lead) is fully submerged equates to the boyant force of \( 0.0065 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.8 = 63.7 \, \text{N} \). Since the weight of wood is 31.85 N, the lead must exert \( 63.7 - 31.85 = 31.85 \, \text{N} \). Solve \( m_{lead} \times g = 31.85 \), giving \( m_{lead} = \frac{31.85}{9.8} \approx 3.25 \, \text{kg} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Specific Gravity
Specific gravity, often abbreviated as SG, is a dimensionless quantity that represents the ratio of the density of a substance to the density of a reference substance. In most cases, for liquids and solids, the reference substance is water at a temperature of 4°C, where it has maximum density. An SG value tells us how heavy a substance is compared to water, without regard to the actual units of measurement.
For example, the specific gravity of wood in our problem is 0.50. This means the wood's density is half that of water. Because the specific gravity is less than 1, the wood will float. However, only half of its volume will be submerged when it is floating naturally in water.
This concept is useful in buoyancy calculations, as it helps predict whether an object will float or sink when placed in a fluid. It acts as a comparative tool, simplifying density-related problems by reducing them to ratios.
Density of Substances
The density (\( \rho \) ) of a substance is a measure of how much mass is packed into a given volume. It is usually expressed in kilograms per cubic meter (kg/m³) or grams per cubic centimeter (g/cm³). The formula for density is \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume.
Understanding the density of a substance allows you to calculate its specific gravity or determine its behavior when immersed in a fluid. In this exercise, the wood has a density of 500 kg/m³, calculated by multiplying its specific gravity (0.50) by the density of water (1000 kg/m³).
Different substances have distinct densities, which influence how they interact with each other in terms of floating and sinking. Knowing the density helps solve buoyancy problems, as it provides insights into how much of a substance will be submerged when in water or any other fluid, and how much force is required to make it entirely sink.
Archimedes' Principle
Archimedes' principle is a fundamental law of physics that describes how buoyancy works. According to this principle, any object fully or partially submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object.
This concept is key to understanding why and how objects float. When you place an object in a fluid, like water, it pushes some of the fluid out of the way. The fluid then pushes back up against the object with a force equivalent to the weight of the displaced fluid.
In the exercise, the wooden block initially floats because the buoyant force perfectly balances the weight of the wood. When the lead is attached, it increases the weight, making it necessary for a larger volume of water to be displaced to maintain buoyancy. Eventually, the additional weight from the lead tips the balance, causing the combined weight of the wood and lead to match the buoyant force from the fully displaced water, leading to submersion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(I) Estimate the pressure needed to raise a column of water to the same height as a \(35-\mathrm{m}\) -tall oak tree.

(II) A cube of side length \(10.0 \mathrm{~cm}\) and made of unknown material floats at the surface between water and oil. The oil has a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\). If the cube floats so that it is \(72 \%\) in the water and \(28 \%\) in the oil, what is the mass of the cube and what is the buoyant force on the cube?

(II) A 6.0-cm-diameter horizontal pipe gradually narrows to \(4.5 \mathrm{~cm} .\) When water flows through this pipe at a certain rate, the gauge pressure in these two sections is \(32.0 \mathrm{kPa}\) and \(24.0 \mathrm{kPa}\), respectively. What is the volume rate of flow?

(III) Water stands at a height \(h\) behind a vertical dam of uniform width \(b .(a)\) Use integration to show that the total force of the water on the dam is \(F=\frac{1}{2} \rho g h^{2} b .\) (b) Show that the torque about the base of the dam due to this force can be considered to act with a lever arm equal to \(h / 3\). (c) For a freestanding concrete dam of uniform thickness \(t\) and height \(h,\) what minimum thickness is needed to prevent overturning? Do you need to add in atmospheric pressure for this last part? Explain.

(III) Thrust of a rocket. (a) Use Bernoulli's equation and the equation of continuity to show that the emission speed of the propelling gases of a rocket is $$v=\sqrt{2\left(P-P_{0}\right) / \rho}$$ where \(\rho\) is the density of the gas, \(P\) is the pressure of the gas inside the rocket, and \(P_{0}\) is atmospheric pressure just outside the exit orifice. Assume that the gas density stays approximately constant, and that the area of the exit orifice, \(A_{0},\) is much smaller than the cross-sectional area, \(A,\) of the inside of the rocket (take it to be a large cylinder). Assume also that the gas speed is not so high that significant turbulence or nonsteady flow sets in. (b) Show that the thrust force on the rocket due to the emitted gases is $$F=2 A_{0}\left(P-P_{0}\right)$$

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.