Chapter 3 Static Equilibrium

1 Chapter 3 Static The Important StuffIn this Chapter we study aspecial caseof the dynamics of rigid objects covered in the lasttwo chapters. It is the (very important!) special case wherethe center of mass of the objecthas no motion and the object is not Conditions for Equilibrium of a Rigid ObjectFor a rigid object which is not movingat allwe have the following conditions: The (vector) sum of the external forces on the rigid object must equal zero: F= 0( )When this condition is satisfied we say that the object is intranslational Equilibrium .(It really only tells us thataCMis zero, but of course that includes the case where the objectis motionless.) The sum of the external torques on the rigid object must equalzero. = 0( )When this condition is satisfied we say that the object is inrotational Equilibrium . (Itreally only tells us that about the given axis is zero, but again that includes the casewhere the object is motionless.)

2 When both and are satisfied we say that the object is instatic all of the problems we will solve in this Chapter are two dimensional problems (inthexyplane), and for these, Eqs. and reduce to Fx= 0 Fy= 0 z= 0( )5556 Chapter 3. Static Two Important Facts for Working Statics Problemsi)The force of gravity acts on all massive objects in our statics problems; its acts on all theindividual mass points of the object. One can show that for the purposes of computing theforces and torques on rigid objects in statics problems we can treat the mass of the entireobject as being concentrated at its center of mass; that is, for an object of massMwe cantreat gravity as exerting a forceMgdownward at the center of mass.(This result depends on the fact that the acceleration of gravity,gis usually constantover the volume of the object. Otherwise it is not true.)ii)While there is only one way to write the conditions for theforceson a rigid objectsumming to zero, we have a choice in the way we write the equation for the total does not specify the choice of theaxisfor calculating the torque.

3 In general it mattersa great deal which axis we pick! But when the sum of torques about any one axis is zeroandthe sum of forces is zero (translational Equilibrium ) then the sum of torques aboutanyaxis will give zero; so for statics problems we are free to pick the most convenient axis forcomputing . Often this will be the point on the object where several unknown forces areacting, so that the resulting set of equations will be simpler to Examples of Rigid Objects in Static EquilibriumStrategy for solving problems in Static Equilibrium : Determineallthe forces that are acting on the rigid body. They will come from the otherobjects with which the body is in contact (supports, walls, floors, weights resting on them)as well as gravity, Draw a diagram and put inallthe information you have about these forces: The pointson the body at which they act, their magnitudes (if known), their directions (if known).

4 Write down the equations for Static Equilibrium . For the torque equation you will have achoice of where to put the axis; in making your choice think ofwhich point would make theresulting equations the simplest. Solve the equations! (That s not physics.. that s math.) If the problem is well posed youwill not have too many or too few equations to find all the Worked Examples of Rigid Objects in Static Equilibrium1. The system in Fig. is in Equilibrium with the string in the center exactlyhorizontal. Find (a) tensionT1, (b) tensionT2, (c) tensionT3and (d) angle .[HRW5 13-23] WORKED EXAMPLES5735oq40 N50 NT1T3T2 Figure :System of masses and strings for Example N50 N(a)(b)35oqFigure :(a) Forces at the left junction of the strings. (b) Forces acting at the right junction of !Fourunknowns (T1,T2,T3and ) to solve for! How will we ever figure this out?We consider the points where the strings meet; the left junction is shown in Fig.

5 (a).Since a string under tension pulls inward along its length with a force given by the stringtension, the forces acting at this point are as this junction in the strings is in Static Equilibrium ,the (vector) sum of the forcesacting on it must give zero. Thus the sum of thexcomponents of the forces is zero: T1sin 35 +T2= 0( )and the sum of theycomponents of the forces is zero:+T1cos 35 40 N = 0( )Now we look at the right junction of the strings; the forces acting here are shown inFig. (b). Again, the sum of thexcomponents of the forces is zero: T2+T3sin = 0( )and the sum of theycomponents of the forces is zero:+T3cos 50 N = 0( )58 Chapter 3. Static EQUILIBRIUMAnd at this point we are done with thephysicsbecause we have four equations for fourunknowns. We will do algebra to solve for this problem the algebra really isn t so bad. From Eq. we getT1=(40 N)(cos 35 )= Nand then Eq. gives usT2:T2=T1sin 35 = ( N) sin 35 = now rewrite Eq.

6 As:T3sin =T2= N( )and Eq. as:T3cos = N( )Now if we divide the left and right sides of by the left and right sides of we get:tan =( N)( N)= then = tan 1( ) = Finally, we getT3from Eq. :T3=( N)(cos )= NSummarizing, we have found:T1= NT2= NT3= N = This answers all the parts of the The system in Fig. is in Equilibrium . A mass of225 kghangs from the endof the uniform strut whose mass kg. Find (a) the tensionTin the cableand the (b) horizontal and (c) vertical force components exerted on the strut bythe hinge.[HRW5 13-33](a)The rigid body here is the strut. What are the forces acting onit?We know the massMof the strut; the force of gravity exerts a forceMgdownward atits center of mass (which is in the middle of the strut since itis uniform). If the hangingmass ism= 225 kg then the string which supports it exerts a downward force of magnitudemgat the top end of the strut.

7 The cable attached to the top of thestrut exerts a forceof magnitudeT. What is its direction? Some geometry (shown in Fig. ) shows that WORKED EXAMPLES5930o45o225 kgTHingeStrutFigure :Geometry of the statics problem of Example :Forces acting on the strut in Example 3. Static Equilibrium direction makes an angle of 15 with the strut. Finally thehingeexerts a force on the strut.(Can t forget that.. the hinge is in contact with the metal bar which is the strut as soexerts a force on it.) The magnitude of this force is just labelledFhin the diagram, but wedon t know its direction!Now, one way to solve the problem would be to let the directionof the hinge force besome angle as measured from some line of reference. In fact it will probably be easiest tolet thexandycomponentsof this force be the I will call themFh,xandFh,y. Infact, parts (b) and (c) of the problem ask us for these components directly.

8 We can alwaysget the direction and magnitude later!Now let s write down some equations. First, the sum of theFx s must give zero. Note(from basic geometry) that the force of the cable is directedat 30 below the force of the hinge has anxcomponent! Then from Fig. we immediately read off:Fh,x Tcos 30 = 0( )Good enough. Now onto theFyequation. The sum of theycomponents of the forcesgives zero, and we write:Fh,y Tsin 30 Mg mg= 0( )Now we use the condition for zero net torque. The question is:Where do we want toput the axis? For this problem, the answer is obvious. We wantto put it at the hingeitself because then when we calculate the torques, the hingeforce (with its two unknowncomponents) will givenotorque. The equation will still be useful.. and it will be muchsimpler. (Keep in mind that even though a physical strut really does turn around a physicalhinge we still have thechoiceof putting the axis for torque anywhere.)

9 We are not told the length of the strut, so let its length beL. We note the angles thatthe force vectors make with the line joining the axis to the points of application, and thenwe write the sum of the torques as: MgL2sin 45 mgLsin 45 +T Lsin 15 = 0but we note that we can cancel theLout of this equation, leaving Mg2sin 45 mgsin 45 +Tsin 15 = 0( )Are we done with thephysicsyet? In the Eqs. , and there are threeunknowns:T,Fh,xandFh,y. Wearedone with the physics. Only algebra the algebra isn t so bad. Theonlyunknown in Eq. isTand we get:Tsin 15 =Mg2sin 45 +mgsin 45 =12(45 kg)( ) sin 45 + (225 kg)( ) sin 45 = 1715 Nso thatT=(1715 N)sin 15 = WORKED EXAMPLES61CM60oLxLL/2mM=2mFigure :Geometry of the statics problem of Example 3. Student is standing on a ladder which leansagainst a wall.(b)With our result from part(a) in hand,Fh,xandFh,ywill be easy to find. From weget:Fh,x=Tcos 30 = (6630 N) cos 30 = 103 Nand From we get:Fh,y=Tsin 30 +Mg+mg= (6630 N) sin 30 + (45 kg)( ) + (225 kg)( )= 103 NThe horizontal and vertical components of the force of the hinge on the strut areFh,x= 5740 NFh,y= 5960 A ladder having a uniform density and a massmrests against a frictionlessvertical wall at an angle of60.

10 The lower end rests on a flat surface where thecoefficient of Static friction is s= A student with a massM= 2mattemptsto climb the ladder. What fraction of the lengthLof the ladder will the studenthave reached when the ladder begins to slip?[Ser4 12-13]We make a basic diagram of the geometry of the problem in Fig. The ladder haslengthL; we show the center of mass of the ladder at a distanceL2up from its bottom the student had climbed a fractionxof the ladder, then he/she is at a distancexLfromits lower end, as the geometry in mind, we next think aboutallthe separate forces that are actingon the ladder as it leans against the wall and supports the force of gravitymg(downward) is effectively exerted at the center of the ladder. Sincethe ladder is exerting an upward forceMgon the student, the student must be exerting a62 Chapter 3. Static EQUILIBRIUMCMNwNfmgfsMgFigure :Forces acting on the ladder in Example 3.