2. 1. The vector sum of all the external forces on the body
must be zero 2. The vector sum of all the external torques 0netF =
2. The vector sum of all the external torques that act on the body
measured about any point must be zero 0net =
3. Statics Problem Recipe 1. Draw a force diagram. (Label the
axes) 2. Choose a convenient origin O. A good choice is to have one
of the unknown forces acting at O 3. Sign of the torque for each
force: - If the force induces clockwise (CW) rotation- + If the
force induces counter-clockwise (CCW) rotation 4. Equilibrium
conditions: 5. Make sure that: numbers of unknowns = number of
equations , , , 0 0 0 net x net y net z F F = = =
4. Auniformbeamof length and mass = 1.8 kg is at rest on two
scales. Auniformblock of mass = 2.7 kg is a rest on the beamat a
distance /4 fr Sample Prob omits left lem end. 12- Calc l 1. u ate
L m M L , the scales readings 0 (eqs.1) We choose to calculate the
torque with repsect to an axis through the left end of the
beam(point O). net y rF F F Mg mg= + = O axis through the left end
of the beam(point O). (12-9)
5. A l a d d e r o f l e n g t h = 1 2 m a n d m a s s = 4 5 k
g l e a n s a g a i n s t a f r i c t i o n l e s s w a l l . T h e
l a d d e r 's u p p e r e n d i s a t a h e i g h t = 9 . 3 m a b
o v e t h e p a v e m e n t o n w h i c h S a m p l e t h e l o w e
r e n d r e s t s . P r o b l e m 1 2 - 2 : L m h 2 2 T h e c o m o
f t h e l a d d e r i s / 3 f r o m t h e l o w e r e n d . A f i r
e f i g h t e r o f m a s s = 7 2 k g c l i m b s h a l f w a y u p
t h e l a d d e r F i n d t h e f o r c e s e x e r t e d o n t h e
l a d d e r b y t h e w a l l a n d t h e p a v e m e n t . D i s t
a n c e 7 . 5 8 L M a L h= = m
6. ( )( ) ( ) ( ) ( ) , Wetaketorquesabout anaxisthroughpoint
O. 0 3 2 9.8 7.58 72/2 45/32 3 407 N 410N 9.3 net z w w a a h F mg
Mg M m ga F h = + + = + + = = = , , 9.3 0 410N 0 9.8 72 4 net x w
px px w net y py py h F F F F F F F Mg mg F Mg mg = = = = = = = + =
+( )5 1146.6N 1100N=
7. A safe of mass = 430 kg hangs by a rope from a boom with
dimensions = 1.9 m and = 2.5 m. The beam of the boom has mass = 85
kg Find the tension Sample in Problem the cabl 12-3: e and the mc M
a b m T ( )( ) ( )( ) ( ), agnitude of the net force exerted on the
beam by the hinge. We calculate the net torque about an axis normal
to the page that passes through point O. 0 2 net z c r F b a T b T
mg = = ( )( ) ( )( ) ( ), 2 9.82 net z c r c m gb M T a + = = (
)2.5 430 85/2 6100 N 1.9 + ( ) ( ) ( ) ( ) , , 2 22 2 0 6093 N 0
9.8 85 430 5047 N 6093 5047 7900 N net x h c h c net y v r v r h v
F F T F T F F mg T F mg T g m M F F F = = = = = = = + = + = + = = +
= +
8. The center of mass of a system of particles is the point
that moves as though; i) All of the systems mass were concentrated
there ii) All external forces were applied there.
9. Example:
10. Thegravitational forceactingonanextendedbodyis thevector
sumof the gravitational forces actingontheindividual elements of
thebody. The Thecenter gravitati of G onal f ravity(c orce g)o gF
onabodyeffectivelyacts at asinglepoint knownas thecenter of gravity
of thebody. Here"effect If theindividual gravitational ively"has
thefollowingmeaning forces ontheelements of th : ebodyareturnedoff
andreplacedby actingat thecenter of gravity, thenthenet
forceandthenet torqueabout anypoint gF actingat thecenter of
gravity, thenthenet forceandthenet torqueabout anypoint
onthebodydoes not chang Weshall provethat if theaccelerationo ge. f
ra center of gravity center of mass vity is thesamefor all
theelements of thebodythenthe coicides with the . This is
areasonableapproximationfor objects near thesurface of
theearthbecause g changes verylittle.g
11. Consider the extended object of mass shown in fig.a. In
fig.a we also show the i-th element of mass . The gravitational
force on is equal to where is the acceleration of gravity in i i i
i i M m m m g g the vicinity of . The torque on is equal to . The
net torque (eqs.1) Consider now fig.b in which we have replaced the
forces by the net gravitational force i i i gi i net i gi i i i gi
g m m F x F x F F = = acting at the center of gravity. The net
torque is equal to : (eqs.2)net net cog g cog gi i x F x F = = i If
we compare equation 1 with equation 2 we get: We substitute for and
we have: If we set for all the elements cog gi gi i i i i i gi cog
i i i i i i i i i i i cog com i i x F F x m g F x m g m g x m x g g
x x m = = = = = If g has the same value at all points on a body,
its center of gravity is identical to its center of mass.
12. In the three figures above we showthe three ways in which a
solid might change its dimensions under the action of external
deforming forces. In fig. the cylinder is stretched byforces acting
alon a g the cylinder axis. In fig. the cylinder isbis stretched
byforces acting along the cylinder axis. In fig. the cylinder is
deformed byforces perpendicular to its axis. In fig. a solid placed
in a fluid under high pressure is compressed uniformlyon all sides.
All three d b c eformation types have in common (defined as
deforming force per unit area). These stresses are k
tensile/compressive shearingnown as for fig.a, for fig.b, and for
fig.c.hydra The st apu p ress lilic cation of stress on a solid
results in strain which take different formfor the three types of
strain. Strain is related to strain via stress =the equa
modulustion s: train
13. is definedas theratio where is thesolidarea is definedas
theratio where is thechangeinthelength of the (sym cyli
Tensilestress Strain ndrical solid. Stre bol ss ) is plottedv F A L
L A LS L ersus strainin theupper figure.Stress is plottedversus
strainin theupper figure. For awiderangeof appliedstresses the
stress-strainrelationis linear andthesolidreturns toits original
lengthwhenthestress is removed. This is knownas the . If thestress
is increasedbeyondamaximumvalue knownas theyieldstrngth thecylinder
becomes permanently deformed. If thestress continues
toincreasethecyl elastic inder br rang ea e ks a yS t astress
valueknownas ultimatestrength uS