Force - Engineering Mechanics - Statics - Lecture Slides, Slides of Mechanical Engineering

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Chp 2: Force

DeComposition

Force Defined

• Force: Action Of One Body On Another ; Characterized By Its - Point Of Application - Magnitude (intensity) - Direction

 The DIRECTION of a Force Defines its Line of Action (LoA)

Magnitude

Line of Action

Direction

Point of Application

Weight

• Consider An Object of mass, m, at a modest Height, h, Above the Surface of the Earth, Which has Radius R - Then the Force on the Object (e.g., Yourself)

m [ ] g R

F m GM R h

F GMm = 

 

⇒ ≈ 

= 2 but R h 2

W = mg

 This Force Exerted by the Earth is called Weight

• While g Varies Somewhat With the Elevation & Location, to a Very Good Approximation - g ≈ 9.81 m/s 2 ≈ 32.2 ft/s 2

Earth Facts

• D ≈ 7 926 miles (12 756 km)
• M ≈ 5.98 x 10^24 kg
  • About 2x10^15 Empire State Buildings
• Density, ρ ≈ 5 520 kg/m^3
  • ρwater ≈ 1 027 kg/m 3
  • ρsteel ≈ 8 000 kg/m 3
  • ρglass ≈ 5 300 kg/m 3

Europa Weight

• Since your MASS is SAME on both Earth and Europa need to Find only geu and compare it to gea
• Recall

R^2

GM

g ≈

 Europa Statistics from table:  M (^) eu = 4.8x10^22 kg  Reu = 1 569 km

 Then g (^) eu

( 3 )^2

22 2 11 3 1569 10

6 673 10 48 10 m

kg kg s g (^) eu m ×

× ≈ × ⋅

. −^.

( )( ) 2 2

3 12

11 22 2 462 10

6673 10 48 10 kg s m g (^) eu m kg ⋅ ⋅

⋅ × ≈ × − × .

..

g 1 301 m s^2 eu ≈.  With %Weu = g (^) eu /g (^) ea

. % .

%. 13 27 9 807

W^1301 eu = =

Contact Forces

• Normal Contact Force
  • When two Bodies Come into Contact the Line of Action is Perpendicular to the Contact Surface

 Friction Force

• a force that resists the relative motion of objects that are in surface contact - Generation of a Friction Force REQUIRES the Presence of a Normal force

Contact Forces

• Compression Force
  • A PUSHING force which tends to SMASH an object upon application of the force

 Shear Force

• a force which acts across a object in a way that causes one part of the structure to slide over an other when it is applied

Recall Free-Body Diagrams

 SPACE DIAGRAM ≡ A Sketch Showing The Physical Conditions Of The Problem

 FREE-BODY DIAGRAM ≡ A Sketch Showing ONLY The Forces On The Selected Body

Vector Notation – Unit Vectors

• Unit Vectors have, by definition a Magnitude of 1 (unit Magnitude)
• Unit vectors may be
  • Aligned with the CoOrd Axes to form a Triad
  • Arbitrarily Oriented

i ≡ i ˆ j ≡ ˆ j k ≡ k ˆ u ≡ u ˆ λ ≡ λ ˆ

 Unit Vectors may be indicated with “Carets”

Example: FBD & Force-Polygon EYE, Not Pulley

 A 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope?

 SOLUTION PLAN:

• Construct a free-body diagram for the rope eye at the junction of the rope and cable. - i.e., Make a FBD for the connection Ring-EYE
• Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the connecting eye.
• Apply trigonometric relations to determine the unknown force magnitudes

Vector Notation – Vector ID

• In Print and Handwriting We Must Distinguish Between - VECTORS - SCALARS
• These are Equivalent Vector Notations

P ≡ P ≡ P ≡ P

• Boldface Preferred for Math Processors
• Over Arrow/Bar Used for Handwriting
• Underline Preferred for Word Processor

Vector Notation - Magnitude

• The Magnitude of a vector is its Intensity or Strength - Vector Mag is analogous to Scalar Absolute Value → Mag is always positive - Abs of Scalar x → | x | - Mag of Vector P → ||P|| =
• We can indicate a Magnitude of a vector by removing all vector indicators; i.e.:

P = P ≡ P ≡ P ≡ P ≡ Mag of P

Angle Notation: Space ≡ Direction

• The Text uses [ α , β , γ ] to denote the Space/Direction Angles
• Another popular Notation set is [ θx , θy , θz ]
• We will consider these Triads as Equivalent Notation: [ α , β , γ ] ≡ [ θx , θy , θz ]

Magnitude-Angle Form

• The Magnitude of the Force is Proportional to the Geometric Length of its vector representation: F ∝ L where L is the Pythagorean Length :

L = ( x 2 − x 1 )^2 + ( y 2 − y 1 )^2 + ( z 2 − z 1 )^2

 Note that if Pt 1 is at the ORIGIN and Pt 2 has CoOrds (x, y, z) then

L = x^2 + y^2 + z^2