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Chapter 4: Problem 38

In The cgs unit of force is the \(d\) yne, equal to \(1 \mathrm{~g} \mathrm{~cm} / \mathrm{s}^{2}\). Find the conversion factor between dynes and newtons.

Short Answer

Expert verified
1 dyne = \(10^{-5}\) newtons.

Step by step solution

01

Understand the Units

First, let's understand the units involved. The dyne is the unit of force in the cgs (centimeter-gram-second) system, where 1 dyne is equal to 1 gram centimeter per second squared.
02

Understand Newtons

In the SI system, the unit of force is the newton. One newton is equal to 1 kilogram meter per second squared.
03

Convert Grams to Kilograms

Since 1 kilogram equals 1000 grams, then 1 gram = 0.001 kilograms.
04

Convert Centimeters to Meters

Since 1 meter equals 100 centimeters, then 1 centimeter = 0.01 meters.
05

Calculate Dyne in Terms of Newtons

Using the conversions:1 dyne = 1 gram \( \times \) 1 centimeter / second虏= 0.001 kg \( \times \) 0.01 m / second虏 = 0.00001 kg m / second虏= \(10^{-5}\) N.

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Key Concepts

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

Understanding the CGS System
The CGS system, which stands for centimeter-gram-second, is one of the several systems used to measure physical quantities. In this system, the primary units are:
  • Centimeter for length
  • Gram for mass
  • Second for time
It is a metric system, like the more commonly used SI system, but it is based on these smaller-sized units. The CGS system is often used in fields such as physics, particularly in areas dealing with electromagnetism and optics. It's important to understand the units within the CGS system as they allow us to grasp the foundational concepts that link to other measurement systems, including the measurement of force, which in CGS is termed a dyne.
Grasping the SI System
The SI system, or the International System of Units, is the most globally accepted method for measuring physical quantities. It uses base units that are:
  • Meter for length
  • Kilogram for mass
  • Second for time
This system helps to maintain consistency and universality in scientific measurements. The SI system's unit of force is the newton, named after Sir Isaac Newton for his important contributions to physics. When discussing the relationships between different units within the SI system, it is essential to note that they are larger than their CGS counterparts, allowing more convenience in expressing everyday measurements.
Deciphering Force Measurement
Force is a vector quantity, which means it has both magnitude and direction. It is measured by how much it can change an object's motion. In both the CGS and SI systems, force is defined using the basic equation:\[ \text{Force} = \text{Mass} \times \text{Acceleration}\]In the CGS system, mass is in grams, and acceleration is in centimeters per second squared, leading to the dyne as the unit of force. Conversely, in the SI system, mass is in kilograms and acceleration in meters per second squared, resulting in the unit of force being the newton. Understanding these distinctions lays the groundwork for understanding how we can convert between these units when necessary.
Simple Dyne to Newton Conversion
Converting between dynes and newtons involves understanding the relationship between CGS and SI system units. The dyne, as per the CGS system, is defined as 1 gram centimeter per second squared. Meanwhile, the newton, in the SI system, is defined as 1 kilogram meter per second squared.Here's how we can perform the conversion:
  • 1 gram equals 0.001 kilograms
  • 1 centimeter equals 0.01 meters
Putting these conversions into practice:
  • 1 dyne = 1 gram 脳 1 cm / second虏
  • = 0.001 kg 脳 0.01 m / second虏
  • = 0.00001 kg m / second虏
  • = \(10^{-5}\) N (newtons)
Thus, 1 dyne equals \(10^{-5}\) newtons. This conversion is crucial for anyone working across systems or striving to understand how various units measure force.

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Most popular questions from this chapter

In constant force \(35.2 \mathrm{~N} \hat{\imath}\) changes a ball's velocity from \(-3.25 \mathrm{~m} / \mathrm{s} \hat{\imath}\) to \(+4.56 \mathrm{~m} / \mathrm{s} \hat{\imath}\) in \(3.50 \mathrm{~s} .\) Find the ball's mass.

A block of mass \(m_{1}=0.560 \mathrm{~kg}\) is placed on top of another block of mass \(m_{2}=0.950 \mathrm{~kg},\) which rests on a frictionless horizontal surface. A horizontal force of \(3.46 \mathrm{~N}\) is applied to the lower block. (a) Draw force diagrams for the upper block and for the two-block system. (b) What minimum coefficient of static friction between the two blocks will prevent the upper one from slipping?

A \(78.0-\mathrm{kg}\) person falls straight down from a \(1.60-\mathrm{m}\) height (meas ured from his feet) and lands with weight distributed equally on both feet. To soften the blow, he bends his knees so that it takes \(0.750 \mathrm{~s}\) for him to stop once his feet touch the ground. (a) What constant force does the floor exert on each foot while he's stopping? (b) Suppose instead that he lands stiff-legged and stops in only \(0.100 \mathrm{~s}\). What force does the floor now exert on each foot? (c) In which case is he more likely to sustain injury? Why?

Experiments show that a walking person's hips describe circular arcs centered on the point of contact with the ground, and having radii equal to the leg's length \(L\) (see Figure \(\mathrm{P} 4.105\) ). Since the person's center of mass (more in Chapter 6 ) is near the hip, we can model the walker as a mass \(M\) moving in a circular arc of radius \(L\). In this case, \(M\) is the mass above the hip, which is roughly the person's total mass. At maximum speed, gravity alone is sufficient to provide the centripetal force. (a) Apply Newton's second law and show that the maximum speed at which a person can walk, according to this model, is \(v_{\max }=\sqrt{L g}\). (To move faster, one must run.) (b) What's the fastest walking speed for a typical \(75-\mathrm{kg}\) adult male? Use measurements on yourself or a friend to determine \(L\).

You're in a car rounding a curve at a constant speed. Is the car's interior an inertial or noninertial reference frame? Explain.
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