e14 - applied mechanics: statics - stanford universitybiomechanics.stanford.edu/e14/e14_s06.pdf ·...

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e14 - applied mechanics: statics

YOU’RE BUSY WITH YOUR 2ND HOMEWORK.

2syllabus

e14 - applied mechanics: statics

second homework

due first midterm:takehome

3homework #02

e14 - applied mechanics: statics

4homework #02

e14 - applied mechanics: statics

textbook.russell c. hibbelerprentice hall, 12th editionengineering mechanics staticsread chapters 3 and 4for homework #2

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63. equilibrium of a particle

• to introduce the concept ofthe free-body diagram for aparticle• to show how to solveparticle equilibrium problemsusing the equations ofequilibrium• when cables are used forhoisting loads, the must beselected so that they do notfail. today, we will show howto calculate cable forces forsuch cases

what we‘ve done so far…

73.1 equilibrium condition of a particle

newton‘s three laws of motion

• first law equilibrium if ∑F = 0 then v = const.

FAB

FBA

• second law accelerated motion F = m · a

• third law actio = reactio FAB = - FBA

= 0

83.2 free body diagram

procedure for drawing a FBD

I. isolate the particle ofinterest - easy ;-)here shown for particle A

II. show all forces - tricky!3 cables, 3 tension forcesassume directions

III. label each force - easy ;-)

93.3 coplanar force systems

conceptual problem 3-1show that the longer the cables,the less the forces in each cable.

FAB FAC

WFAB FAC

W

2!

xy

for longer cables, the angle !becomes smaller, cos! becomesbigger, and FAB and FAC becomesmaller.

W

FAB

FAC

W

FAB

FAC

! !

104. force system resultants

• to discuss the concept of amoment of a force• to calculate the moment ofa force in 2d and 3d using thescalar formulation• to discuss the cross productas a tool to calculate moments• to calculate the moment ofa force in 2d and 3d using thevector formulation

today‘s objectives

114.1 moment - scalar formulation

engineering intuition

largest moment smaller moment no moment

M0 = F · d [M] = N!m = lb!ft

O … fixed pointF … forced … moment arm

124.1 moment - scalar formulation

example 4.1determine the moment M0 of the force F about point O

use positive M0 !+

134.1 moment - scalar formulation

example 4.2

144.2 cross product

right-handed system

C = A ! B = A · B · sin! uC

C = A ! B = det Ay By yAx Bx x

Az Bz z

C = Az Bx - Ax Bz

Ay Bz - Az By

Ax By - Ay Bx

• C is a vector• C is orthogonal to A and B• C is the area enclosed by A and B, i.e., A · B · sin!

+

154.2 cross product

right-handed system

y

z

… if your thumb is the x axis…

… and your middle finger is the z axis…

x

… then your index finger is…

164.2 cross product

right-handed system

y

z

x

… pointing right at your abs …

174.3 moment - vector formulation

right-handed system

MO = r ! F = r · F · sin! uM

MO = r ! F = det ry Fy yrx Fx x

rz Fz z

MO = rz Fx - rx Fz

ry Fz - rz Fy

rx Fy - ry Fx

• MO is a vector• MO is orthogonal to r and F• MO is the area enclosed by r and F, i.e., r · F · sin!

+

+

184.4 principle of moments

example 4.5

determine the moment M0 of the force F about point O

use two different approaches and compare the results!

+

194.4 principle of moments

example 4.5

determine the moment M0 of the force F about point O+

204.4 principle of moments

example 4.5

determine the moment M0 of the force F about point O+