Forces in TranslationalEquilibrium

Say Thanks to the AuthorsClick http://www.ck12.org/saythanks

(No sign in required)

To access a customizable version of this book, as well as otherinteractive content, visit www.ck12.org

CK-12 Foundation is a non-profit organization with a mission toreduce the cost of textbook materials for the K-12 market bothin the U.S. and worldwide. Using an open-content, web-basedcollaborative model termed the FlexBook®, CK-12 intends topioneer the generation and distribution of high-quality educationalcontent that will serve both as core text as well as provide anadaptive environment for learning, powered through the FlexBookPlatform®.

Copyright © 2013 CK-12 Foundation, www.ck12.org

The names “CK-12” and “CK12” and associated logos and theterms “FlexBook®” and “FlexBook Platform®” (collectively“CK-12 Marks”) are trademarks and service marks of CK-12Foundation and are protected by federal, state, and internationallaws.

Any form of reproduction of this book in any format or medium,in whole or in sections must include the referral attribution linkhttp://www.ck12.org/saythanks (placed in a visible location) inaddition to the following terms.

Except as otherwise noted, all CK-12 Content (includingCK-12 Curriculum Material) is made available to Usersin accordance with the Creative Commons Attribution/Non-Commercial/Share Alike 3.0 Unported (CC BY-NC-SA) License(http://creativecommons.org/licenses/by-nc-sa/3.0/), as amendedand updated by Creative Commons from time to time (the “CCLicense”), which is incorporated herein by this reference.

Complete terms can be found at http://www.ck12.org/terms.

Printed: June 28, 2013

www.ck12.org Concept 1. Forces in Translational Equilibrium

CONCEPT 1 Forces in TranslationalEquilibrium

Objective

The student will:

• Understand how to apply Newton’s Second Law under equilibrium conditions in two dimensions.

Vocabulary

• static equilibrium: Objects in static equilibrium are motionless.

Introduction

FIGURE 1.1

There are many objects we do not want to see in motion. In the Figure 1.1, the mountain climbers want their ropesto keep them from moving downward. We construct buildings and bridges to be as motionless as possible. Wewant the acceleration (and velocity) of these objects to be zero. For an object to be in static equilibrium (that is,motionless) the right-hand side of Newton’s Second Law, ∑F = ma, must be zero. Thus, ∑F = 0. This equation issimple enough when an object is held with a single support. In an earlier example, we depicted Joe Loose hangingby a single rope. Joe’s goal was to remain hanging in equilibrium (just like the climbers in the photograph). Theforce of gravity pulling Joe down was exactly balanced by the tension in the rope that supported him.

But Joe won’t be hanging for very long, will he? You can see that the rope is slowly fraying against the mountainside(recall the original problem). Soon it will snap. But Joe’s in luck, because a rescue team has come to his aid. Theyarrive just in time to secure two more ropes to the mountain side and toss Joe the slack to tie around his waist beforehis rope snaps! Joe is saved. But does Joe thank the rescue team like any sane person would? No. Instead, stillin midair, he pulls out a pad and pencil from his back pocket in order to analyze the forces acting on him (Figurebelow).

1

www.ck12.org

FIGURE 1.2

Static Equilibrium—Saving Joe

In order for Joe to remain in equilibrium, he must not move in the x− or y−directions. This means that the sum ofall forces in the x−direction must add to zero. And the sum of all forces in the y−direction must add to zero.

The procedure for solving problems with forces in equilibrium is as follows:

1. Place Figure above in a coordinate plane with the object at the origin.2. Resolve the tension vectors T1 and T2 into their x− and y−components.3. Use Newton’s Second Law: ∑Fx = 0 and ∑Fy = 0.

In order to solve this problem, we’ll need more information, including the angles that the ropes make with thevertical. The information is provided below, along with Figure below.

2

www.ck12.org Concept 1. Forces in Translational Equilibrium

FIGURE 1.3

θ1 = 45◦

θ2 = 30◦

Mg = 800 N

Find T1 and T2

The solution requires solving a set of simultaneous equations.

First, we find the components of vectors T1 and T2.

T1x = T1 sin45◦ and T1y = T1 cos45◦

T2x = T2 sin30◦ and T2y = T2 cos30◦

Next we apply Newton’s Second Law.

∑Fx : T2 sin30◦−T1 sin45◦ = 0

∑Fy : T2 cos30◦−T1 cos45◦ = 800

The first equation can be quickly simplified to give T2 =√

2T1. T2 is then substituted in the second equation and T1is found. Once T1 is found, T2 can easily be computed using T2 =

√2T1.

3

www.ck12.org

FIGURE 1.4

T1 = 414.11 = 414 = 410 N

T2 = 585.56 = 586 = 590 N

Check your understanding

What general equation can be written for ∑y

F if the angles in Figure above are equal?

Answer: The sum of the y−components is responsible for supporting the weight. If the angles are equal, the ropeshave equal tension. Therefore, the tension in the y−direction of either rope can quickly be found since:

∑y

F = 2Ty = mg→ Ty =mg2

4

www.ck12.org Concept 1. Forces in Translational Equilibrium

References

1. Image copyright Gregory Epperson, 2012. http://www.shutterstock.com. Used under license from Shutter-stock.com

2. Image copyright artenot, 2012; modified by CK-12 Foundation - Raymond Chou. http://www.shutterstock.com. Used under license from Shutterstock.com

3. Image copyright artenot, 2012; modified by CK-12 Foundation - Raymond Chou. http://www.shutterstock.com. Used under license from Shutterstock.com

4. Image copyright artenot, 2012; modified by CK-12 Foundation - Raymond Chou. http://www.shutterstock.com. Used under license from Shutterstock.com

5