1. 1. EQUILIBRIUM
2. 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. 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. 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. 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. 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. 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. 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. 9. Example:
10. 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. 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. 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. 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
14. 14. A
15. 15. A F