Lecture 3. Mechanical Force

8/17/2019 Lecture 3. Mechanical Force

1/33

Mechanical ForceMechanical Force

Equilibrium

Prof. Dr hab. Zbigniew DunajskiProf. Dr hab. Zbigniew Dunajski


2/33

Translational Equilibrium First Condition of Equilibrium

The net eternal force must be !ero

This is a necessar"# but not sufficient#

condition to ensure that an object is incom$lete mechanical equilibrium This is a statement of translational

equilibrium


3/33

%ector


4/33

Module


5/33

&ddition


6/33

Product of %ectors

Work = force x distance


7/33

Friction

The kinetic frictional force is exerted on the upper body by the

stationary lower body. The upper body is moving with velocity

and is pressed together with the lower body by a normal force . It

may also be acted upon by an additional non-normal external force

Fext .

is called the coefficient of

kinetic friction.

is thecoefficient of static friction. Generally we find that


8/33

TThe larger the force is, the larger thehe larger the force is, the larger the

accelerationacceleration


9/33

Torque

The door is free to rotate about an ais through '

There are three factors that determine theeffecti%eness of the force in o$ening the door( The magnitude of the force The position of the a$$lication of the force The angle at which the force is a$$lied


10/33

Torque# contTorque# cont

Torque# τ# is the tendenc" of aforce to rotate an object about

some ais τ ) r F

τ is the torque F is the force

s"mbol is the *reek tau r is the length of the $osition %ector

+, unit is -.m


11/33

Lever

- F r 1 + Fc r 2 + R.

0 = 0


12/33

e%er &rm


13/33

/ight 0and /ule Point the fingers

in the direction of

the $osition %ector Curl the fingerstoward the force%ector

The thumb $ointsin the direction ofthe torque


14/33

Eam$lea. & man a$$lies a force

as shown. Find the

torque on the doorrelati%e to the hinges.

b. +u$$ose a wedge is$laced 1.23 m from

the hinges. 4hatforce must the wedgeeert so that the doorwill not o$en.


15/33

*eneral Definition of

Torque Taking the angle into account leads to a

more general definition of torque(

τ = r F sin θ F is the force r is the $osition %ector θ is the angle between the force and the

$osition %ector


16/33

Equilibrium Eam$le


17/33

&is of /otation ,f the object is in equilibrium# it does not

matter where "ou $ut the ais ofrotation! for calculating the net torque 'ften the nature of the $roblem will suggest a

con%enient location for the ais 5usuall" toeliminate a torque6

4hen sol%ing a $roblem# "ou must s$ecif" an

ais of rotation 'nce "ou ha%e chosen an ais# "ou must maintain

that choice throughout the $roblem

The fulcrum does matter# but the origin! selected for le%er arms will not


18/33

/otational Equilibrium To ensure mechanical equilibrium#

"ou need to ensure rotational

equilibrium as well as translational The +econd Condition of

Equilibrium states The net eternal torque must be !ero


19/33

Center of *ra%it" The force of gra%it" acting on an

object must be considered

,n finding the torque $roduced b"the force of gra%it"# all of theweight of the object can be

considered to be concentrated at asingle $oint# the center of gra%it"


20/33

Calculating the Center of

*ra%it" The object is

di%ided u$ into alarge number of%er" small $articlesof weight 5mi g6

Each $article will

ha%e a set ofcoordinatesindicating itslocation 5#"6


21/33

Find "our center of gra%it"

Consider a $erson with L ) 178 cm and

weight w ) 712 -. a"ing on a board withweight w b ) 9: -# a scale has a force

reading of F ) 823 -. Find the $erson"scenter of gra%it".


22/33

Eam$le of a Free ;od"

Diagram 5Forearm6

,solate the object to be anal"!ed

Draw the free bod" diagram for that object ,nclude all the eternal forces acting on the object


23/33

-ewton"s +econd aw for a

/otating 'bject

The angular acceleration is directl"$ro$ortional to the net torque

The angular acceleration isin%ersel" $ro$ortional to themoment of inertia of the object

H k ’ L


24/33

Hooke’s Law

k ’


25/33

HookeHooke’’s Laws Law

Force !"

#xtension m"

$radient gives

the spring

constant


26/33

+tress# +train < 0ooke=s aw


27/33

Takes into accountthe thickness andlength of a wire.

>oung Modulus E )stress ? strain.

@nits are - mAB or Pa E can be worked out

from the gradient ofthe stressAstraingra$h.

%tress

&a"

%train

!ot always

easy to see

Elastic

limit

Ultimate

tensile

stress

%naps

Young ModulusYoung Modulus


28/33

Fig '.()* p.))'

%lide (+


Diagram 5;eam6 The free bod"

diagram includes

the directions ofthe forces

The weights actthrough the

centers of gra%it"of their objects


29/33


Diagram 5adder6

The free bod" diagram shows the normal force

and the force of static friction acting on theladder at the ground The last diagram shows the le%er arms for the

forces


30/33

Total Energ" of a +"stem Conser%ation of Mechanical Energ"

/emember# this is for conser%ati%eforces# no dissi$ati%e forces such as

friction can be $resent Potential energies of an" other

conser%ati%e forces could be added


31/33

Energ" Methods & ball of mass M

and radius /

starts from rest.Determine itslinear s$eed atthe bottom of the

incline# assumingit rolls withoutsli$$ing.


32/33

More &ngular Momentum ,f the net torque is !ero# the angular

momentum remains constant

Conservation of Angular Momentum states( The angular momentum of as"stem is conser%ed when the neteternal torque acting on the s"stems is

!ero. That is# when


33/33

Conser%ation of &ngular

Momentum# Eam$le 4ith hands and

feet drawn closer

to the bod"# theskater"s angulars$eed increases is conser%ed# ,

decreases# ω increases

Documents

Lecture 3. Mechanical Force