System isolation and equilibrium conditions

System isolation and equilibrium conditions

Objectives Students must be able to #1
Course Objective State the conditions of equilibrium, draw free body diagrams (FBDs), analyse and solve problems involving rigid bodies in equilibrium. Chapter Objectives Analyse objects (particles and rigid bodies) in equilibriums Classify problems in equilibrium into SD and SI categories

Equilibrium Definition
stationary An object is in equilibrium when it is stationary or in steady translation relative to an “inertial reference frame”. moving with Constant velocity Whether object in stationary (moving in steady translation) or not, depends on “reference frame”. really equilibrium? Equilibrium Centrifugal acceleration Newtonian Mechanics “Inertial Reference Frame” - Earth Frame - Central Universe Frame Earth At universe

3/1 Introduction A - Equilibrium is the most important subject in statics. In statics, we deal primarily with bodies at rest. (i.e. they are in the state of “equilibrium”). A - More precisely, when a body is in equilibrium, the wrench resultant of all forces acting on it is zero; i.e. These requirements are necessary and sufficient conditions for equilibrium; i.e. If is true body in equilibrium If body in equilibrium is true From the Newton’s second law of motion, a body that moves with constant velocity, rotates with constant angular velocity; i.e. “zero acceleration”, can also be treated as in a state of equilibrium.

Equilibrium in 2D 3/2 Mechanical system isolation Isolate body
- All physical bodies are inherently 3D, but many may be treated as 2D; e.g. when all forces are on the same plane. 3/2 Mechanical system isolation Before we apply the equilibrium conditions we need to know what force or couple are involved. Isolate body Draw Free Body Diagram (FBD) FBD is used to isolate body (or bodies / system of bodies) so that force/couple acting on it can be identified.

Free Body Diagram (FBD)
FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings. The isolation of body clearly separate cause and effects of loads on the body. A thorough understanding of FBD is most vital for solving problems.

Construction of FBD y x y x body in interest
1) Pick body/combination of bodies to be isolated 2) Isolating the body. Draw “complete external boundary” of the isolated body Free Body Diagram 3) Add all forces and moments (including that are applied by the removed surrounding) y x 150 cm 50 cm 4) Indicate a coordinate system 5) Indicate necessary dimensions Most important step is solving problems in mechanics. *** If an FBD is not drawn (when it is needed), you will get no credit ( 0 point ) for the whole problem!!!! *** 150 cm y x 50 cm

Note on drawing FBD y x Free Body Diagram
Establish the x, y, z axes in any suitable orientation. Label all the known and unknown force magnitudes and directions on the diagram The sense of a force having an unknown magnitude can be assumed. y x 150 cm 50 cm Use different colours in diagrams Body outline blue Load (force and couple) - red Miscellaneous (dimension, angle, etc.) - black

Equilibrium Solving Procedure
Formulate problems from physical situations. (Simplify problems by making appropriate assumptions) Draw the free body diagram (FBD) of objects under consideration State the condition of equilibrium Substitute variables from the FBD into the equilibrium equations Substitute the numbers and solve for solutions Delay substitute numbers Use appropriate significant figures Technical judgment and engineering sense Try to predict the answers Is the answer reasonable?

Equilibrium Free Body Diagram (FBD)
FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings. The isolation of body clearly separate cause and effects of loads on the body. A thorough understanding of FBD is most vital for solving problems.

Equilibrium FBD Construction
Select the body to be isolated Draw boundary of isolated body, excluding supports Indicate a coordinate system by drawing axes Add all applied loads (forces and couples) on the isolated body. Add all to support reactions (forces and couples) represent the supports that were removed. Beware of loads or support reactions with specific directions due to physical meanings Add dimensions and other information that are required in the equilibrium equation

Equilibrium Help on FBD
Establish the x, y axes in any suitable orientation. Label all the known and unknown applied load and support reaction magnitudes Beware of loads or support reactions with specific directions due to physical meanings Otherwise, directions of unknown loads and support reactions can be assumed.

Equilibrium On FBD Analyses
Objective: To find support reactions Apply the equations of equilibrium Load components are positive if they are directed along a positive direction, and vice versa It is possible to assume positive directions for unknown forces and moments. If the solution yields a negative result, the actual load direction is opposite of that shown in the FBD.

Contents Equilibrium of Objects Particles (2D & 3D) 2D Rigid Bodies
the heart () of Statics Equilibrium of Objects Particles (2D & 3D) 2D Rigid Bodies 3D Rigid Bodies Rigid Bodies SD and SI Problems Particle Ideal particle can not rotate. (no couple acting on it)

Particles FBD construction
To construct a complete FBD of a particle Select the particle to be isolated Draw the particle as a point Indicate a coordinate system Add all active forces/moments (weight, etc.) Add all support reactions (e.g. tension in the tangential direction of a cable, tensile and compressive forces in a compressed and stretched springs)

Particles Equilibrium Analyses #2
Equations of Equilibrium Apply the equations of equilibrium Components are positive if they are directed along a positive axis, and negative if they are directed along a negative axis. Assume the directions of unknown forces in the positive x, and y axes. If the solution yields a negative result, this indicates the sense of the force is the reverse of that shown on the FBD.

Particles Equilibrium in 3D
Particle Equilibrium Particles Equilibrium in 3D z y x

H-Ex3-1#1) The sphere has a mass of 6 kg and is supported as shown
H-Ex3-1#1) The sphere has a mass of 6 kg and is supported as shown.Draw a free-body diagram of the sphere, the cord CE, and the knot at C.

y x How many unknowns? How many Equations? Action-reaction pair
- use same symbol - opposite direction

Determine the tension in cables AB and AD for equilibrium of the 250-kg engine shown.
FBD of A

FBD of A

Example Hibbeler Ex 3-3 #1 2unknown, 2 Eqs (at this stage) If the sack at A has a weight of 20 lb, determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown. 3unknown, 2 Eqs (at this stage) = ? FBD of E

Recommended FBD of E FBD of C

Example Hibbeler Ex 3-5 #1 A 90-N load is suspended from the hook. The load is supported by two cables and a spring having a stiffness k = 500 N/m. Determine the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x-y plane and cable AC lies in the x-z plane.

3D particle Equilibrium
How many unknowns, how many equations? (no FBD, no score) perfect answer sheet

Example Hibbeler Ex 3-7 #1 Determine the force developed in each cable used to support the 40-kN (4 tonne) crate shown. 3D particle Equilibrium

Particle Equilibrium Hibbeler Ex 3-7 #2

Particle Equilibrium Example Hibbeler Ex 3-7 #4

Equilibrium of 2D Rigid Bodies
2D Equilibrium Equilibrium of 2D Rigid Bodies Use similar analyses as the particles Additional consideration Action forces in supports/constraints Free-body diagram (FBD) of 2D rigid bodies Equilibrium equations (scalar form) for rigid bodies Two-force and three-force members

Force Reaction (2D) To write an FBD, first, you will need to know what kind of force we will get when eliminating the environment/surrounding. 1) Flexible cable, belt, chain, or rope   always away from the body T  T tangent to the cable

Not always this direction only this direction
2) Smooth surfaces - Contact force at contact point normal to the surface/contact plane only this direction N - always compressive 3) Rough surfaces Not always this direction A rough surface can produce a tangential force (F, friction) as well as a normal force (N) direction of F depend on situations (chapter 6) N F R only this direction

M existence due to its bending resistance
4) Roller supports N - Roller, rocker, or ball transmits a compressive force normal to the supporting surface only this direction N 5) Freely sliding guide The vector N may be up or down depend on problem. If not known, you may assume any of the two. After further calculation, if N is +, correct sense was assumed. If negative, N goes the other way. N not always this direction N M M existence due to its bending resistance

6) Pin connection Pin free to turn Pin not free to turn Rx Ry Rx Ry M not always this direction As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple is exerted on the body. 7) Built-in or fixed support A not always this direction M F V Weld A A

m G W=mg x F F 8) Gravitational attraction
Resultant of the gravitational attraction is the weight W = mg and act toward center of the earth passing through the center mass G m G W=mg 9) Spring action F x is positive x is negative x F Normal distance For linear springs , F = kx

You may assume either case.
Equilibrium You may assume either case. The sign will indicate its sense of direction later.

Equilibrium construction of a FBD
2D Equilibrium Equilibrium construction of a FBD To construct a complete FBD for a 2D rigid bodies determine which body is to be isolated draw external boundary of isolated body indicate a coordinate system (axes) add all loads (forces and couples, be they applied or support) No FBDs  Cannot apply equilibrium conditions  NO SCORES

V F y x y x V F y x V F Correct? F1 F2 F3 F1 F2 F3 mass m A M P M
W=mg mass m A M P M y x mass m P W=mg A N N Correct? B V F P P y x No internal force is shown in FBD T m A not always “mg” B V F N **Write FBD before allowing any movements of vectors (sliding/free)**

3/A Correct the incomplete FBD
y x for all figures below F Ax Ay

3/B Correct the wrong or incomplete FBD
y x 3/B Correct the wrong or incomplete FBD for all figures below f mg If rough surface F Ox

3/B Correct the wrong or incomplete FBD
y x 3/B Correct the wrong or incomplete FBD for all figures below D C Ax C

We will choose problem 1 and 4 as samples
Equilibrium System Isolation & FBD

A y x mg By f M N m0g |_ with the incline mg Ax Ay y x

2D Equilibrium Example Hibbeler Ex 5-1 #1 Draw the free-body diagram of the uniform beam shown. The beam has a mass of 100 kg.

Example Hibbeler Ex 5-1 #2 2D Equilibrium

2D Equilibrium Example Hibbeler Ex 5-3 #1 Two smooth pipes, each having a mass of 300 kg, are supported by the forks of the tractor. Draw the FBDs for each pipe and both pipes together.

2D Equilibrium

2D Equilibrium Example Hibbeler Ex 5-4 #1 Draw the FBD of the unloaded platform that is suspended off the edge of the oil rig shown. The platform has a mass of 200 kg.

3/3 Equilibrium Conditions
We can find the wrench resultant force P P “equilibrium” “equilibrium” P Equilibrium conditions (3D) of the body is equivalents to (for some point P ) We can prove this! (for all point O) - For co-planer forces only (2D), eq conditions is equivalents to Fx= Fy = Mo = (for all point O) (2D) Fx= Fy = Mo = (for some point O) (2D)

P 3/3 Equilibrium Conditions We can find the wrench resultant force P P P if in “equilibrium” O Equilibrium conditions (3D) of the body is equivalents to ( moment for some specific point P ) - For co-planer forces only (2D), eq conditions is equivalents to Fx= Fy = Mo = (for all point O) (2D) We can prove this! Fx= Fy = Mo = (for some point O) (2D)

We can find the resultant force and couple at any point in the body Easy - Why? No force to cause moment effect O O Can couple exist? Equilibrium conditions (3D) of the body is equivalents to ( moment for all point ) - For co-planer forces only (2D), eq conditions is equivalents to Fx= Fy = Mo = (for all point O) (2D) (trivially easy) We can prove this (later) ! Fx= Fy = Mo = (for some point O) (2D)

Equilibrium and its Independent Equations
(no tendency to initiate translation, no tendency to initiate rotation) A A equilibrium A A any O (trivially easy) (some point: A) (any point: O) Important Meaning A 2D A Only at most 3 independent Equations If is true body in equilibrium If body in equilibrium is true A From the Newton’s second law of motion, a body that moves with constant velocity, rotates with constant angular velocity; i.e. “zero acceleration”, can also be treated as in a state of equilibrium. A

Equilibrium Eqn. for 2D Rigid Bodies
2D Equilibrium Equilibrium Eqn. for 2D Rigid Bodies Scalar Form Particle Rigid Body The sum of the moment about any point O is zero.

2D Equilibrium Procedure #1
Free-Body Diagram Establish the x, y axes in any suitable orientation. Draw an outlined shape of the body. Show all the forces and couple moments acting on the body. Label all the loadings and specify their directions relative to the x, y axes. The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed. Indicate the dimensions of the body necessary for computing the moments of forces.

Equations of Equilibrium When applying the force equilibrium equations, orient the x and y axes along lines that will provide the simplest resolution of the forces into their x and y components. If the solution of the equilibrium equations yields a negative scalar for a force or couple moment magnitude, this indicate that the sense is opposite to that which was assumed on the FBD.

Equations of Equilibrium Apply the moment equation of equilibrium about a point O that lies at the intersection of the lines of action of two unknown forces. In this way, the moments of these unknowns are zero about O, and a direct solution for the third unknown can be determined.

H5.7) Find the tension in the cord at C and the horizontal and vertical component of reactions at pin A. The pulley is frictionless. The body is in equilibrium. Fx = 0 Fy = 0 MAnyPoint = 0 T are the same at both side? + If couple also applies at pully,T are the same in both side? If pulley rotates, T are the same in both side ? with constant angular velocity?

Independent Equations
y ** Whatever 2D-equilibrium conditions you used, at most 3 unknowns (scalar) may be found. x Body in equilibrium Hey , why you cannot get anymore, you can take another points to take moment, and get new equation? not independent equations! Fx= 0 Fy = 0 Mo = 0 (for some point O) ** However, sometimes you can not get those 3 unknowns in 2D problem. Fx= 0 Fy = 0 Mo = 0 (for all point O)

The body is in equilibrium.
H5.7) Find the tension in the cord at C and the horizontal and vertical component of reactions at pin A. The pulley is frictionless. The body is in equilibrium. Fx = 0 Fy = 0 MAnyPoint = 0 Not independent Equations! +

2D Equilibrium Example Hibbeler Ex 5-6 #1 Determine the horizontal and vertical components of reaction for the beam loaded as shown. Neglect the weight of the beam in the calculation.

2D Equilibrium Example Hibbeler Ex 5-6 #2 Find: Ay, Bx, By FBD of ADB

2D Equilibrium Example Hibbeler Ex 5-8 #1 The link shown is pin-connected at A and rests against a smooth support at B. Compute the horizontal and vertical components of reactions at the pin A.

2D Equilibrium Example Hibbeler Ex 5-8 #2 FBD of link BA

tension is internal force.
3/ kg beam has a C.G. at G. The 80-kg man is exerted a 300-N force on the rope. Calculate the force reaction at A which is a weld pin. 3 unknowns, 3 Equations. You can solve it. x y + mg = 80(9.81) 300 N T mg 300 N N =mg? 200(9.81) C N 300N T = 300N? T=300N 80(9.81) 80(9.81) 200(9.81) 4 unknowns, 3 Eq. You can’t solve it? 200(9.81) 300 N You can solve it ! Use FBD of rod T =300N N is internal force. tension is internal force. C

tension is internal force.
3/ kg beam has a C.G. at G. The 80-kg man is exerted a 300-N force on the rope. Calculate the force reaction at A which is a weld pin. 80(9.81) 80(9.81) Sun’s Gravitational force 200(9.81) tension is internal force. C

Draw the FBD of the foot lever shown.
why?

Two-Force Member In Equilibrium
Fx= Fy = Mo = 0 Two-Force Member In Equilibrium To keep the body in equillibrium, the second force must … Body in Equilibrium - be co-linear - direction of opposite side. body in equilibrium - has the same manitude bodies in equilibrium

(with proper moment arm)
Fx= Fy = Mo = 0 Three-Force Member for Equilibrium To keep the body in equillibrium, the second force must … 1) (co-planer) 2) Concurrent OR Parallel (with proper moment arm) concurrent Parallel (with proper moment arm) 4 forces need to be concurrent or parallel to make body in equilibrium? O M = 0

Special Member 2-Force Member #1
2D Equilibrium Special Member 2-Force Member #1 Definition The object is a two-force member when subjected to two equivalent forces acting at different points.

2D Equilibrium Special Member 2-Force Member #2 If a two-force member is in equilibrium, the two forces: are equal in magnitude. are opposite in direction. have the same line of action. There are no couple moments. The ability to recognize 2-force members is important in the analyses of structures.

2D Equilibrium Special Member 2-Force Member #3 Example cable with negligible weight light bar with pin joints and no load

2D Equilibrium Example Special Member 2-Force Member #4

2D Equilibrium Special Member 3-Force Member #1 Definition The object is a three-force member when subjected to three equivalent forces acting at different points. If a three force member is in equilibrium, the three forces are: coplanar and either parallel or concurrent.

2D Equilibrium Special Member 3-Force Member #2 Concurrent Forces Parallel Forces

Draw the FBD of the foot lever shown.
why? Two force in Equilibrium You should write the FBD of 2-force (in equilibrium) like this.

The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A. 4 unknowns, 3 eq. Hibbeler Ex 5-13 x y + 2 unknowns, 3 eq. Two Force Members

O Solve by using 3-force member Concept Three Forces are concurrent. +
Hibbeler Ex 5-13 O x y + 0.5 m =0.7 m Assume Directions 2 force members =0.4 m 3-Unknowns, 3-Eqs

The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A. 2 force members 0.5 m x y + O Ans

Any two-force members or three-force members here?

3/27 In a procedure to evaluate the strength of the triceps muscle, a person pushed down on a load cell with the palm of his hand as indicated in the figure. If the load-cell reading is 160 N, determine the vertical tensile force F generated by the triceps muscle. The mass of the lower arm is 1.5 kg with mass center at G.

y x T Ox 160N Oy 1.5(9.81) 3 Unknown, 3 Equations. You can solve it.
Free Body Diagram If what you want o know is Oy ? T If what you want o know is Ox ? Ox Oy 160N 0.15 0.15 0.25 1.5(9.81) 3 Unknown, 3 Equations. You can solve it. By easier way. Because what we want to know is the force T of triceps. Taking summation of moments at point O will eliminate Ox and Oy which are unknowns out, and the calculation will be easier. + Ans

Technique: Finding unknown force only in 1 step
Concrete slab with mass of 25,000 kg is pulled by cable of tension P. Determine the tension T in the horizontal cable using only 1 equilibrium equation. P x y Given:  = 60° D W = mg A R T 6 m Key Here we have 3 unknowns (P,R,T). To use only 1equilibrium eq. To determine T, we need to take moment at the point where the other unknown forces (R,P) passes.

4 m + P D W = mg A R T 6 m x y ANS

Alternative Equilibrium equations
Recall: In general, you have these three equilibrium equations: (for some point O) Fx= Fy = Mo = (E) - Alternatively, you may use either MA = 0 MB = 0 Fx= 0 MA = 0 MB = 0 Mc = 0 line AB not |_ to the x direction A B C not on the same straight line R=0 R=0 not |_ B B x A C A

400 N 500 N 80 N-m Calculate the magnitude of the force at pin A. Ax
Ay Free Body Diagram By 500 N x y 80 N-m D How many unknowns ? We have 3 unknowns of Ax Ay By

ANS 400 N 500 N 80 N-m Calculate the magnitude of the force at pin A.
Ax Ay Free Body Diagram By 500 N we can determine Ay from + x y 80 N-m we can determine Ax from + Thus, the resultant of force at pin A: ANS

Constraints and Statical determinacy
** need to know when a problem can be solved or what force/couple can be found. Statically Determinate Improper Constraint - Not enough constraint Unknown 3 , Independent eq 2 Unknown 3 , Independent eq 3 2) Statically Indeterminate Not in equilibrium - can’t get all unknowns No solution: (cant maintain moment) Unknown 4 , Independent eq 3 unknown 3 , independent 2

Independent Equations
y ** Whatever 2D-equilibrium conditions you used, at most 3 unknowns (scalar) may be found. x Body in equilibrium Hey , why you cannot get anymore, you can take another points to take moment, and get new equation? not independent equations! Fx= 0 Fy = 0 Mo = 0 (for some point O) ** However, sometimes you can not get those 3 unknowns in 2D problem. Fx= 0 Fy = 0 Mo = 0 (for all point O) Independent Equations on categories of 2D problem. Equilibrium 1) Co-linear x Depend on x-axis selection Choose this direction as x axis Fx= Fy = Mo = 0 (Choose point O on the line of action)

Depend on x-axis selection
2) Concurrent at a point Fx = 0 Fy = 0 Mo = 0 y x concurrent at point O O 3) Parallel Fx = 0 Fy = 0 Mo = 0 Depend on x-axis selection y x 4) General Fx = 0 Fy = 0 Mz = 0 y x

x Note on Solving Problems
1. If we have more unknowns than the number of independent equations, then we have a statically indeterminate situation. We cannot solve these problems using just statics. 2. The point (axis) which we check for the moment, affects the simplicity of the solution. Choosing appropriate one is the key for fast problem solving. The order in which we apply equations may affect the simplicity of the solution. For example, solving  FX = O first allows us to find the first unknown quickly. x 4. If the answer for an unknown comes out as negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem. N If you get N<0, something wrong!

3/48 The small crane is mounted on one side of the bed of a pickup truck. For the position q=45, determine the magnitude of the force supported by the pin at O and the oil pressure P against the 50-mm-diameter piston of the hydraulic cylinder BC. Body in interest: AOBC? AOC? Because We also want to find force P exerted by BC to the object AOC

y FBD x W Ox C Oy 0.34 C O E F 0.36 B F 0.11 G D H Ans To find C: +
q=45 Find P ,O Ignore mass of link CB y FBD x C W Ox Oy C B which dirction? 2 Force Member // with CB Direction // CB Can you find a ? F To find C: 0.11 + 0.34 q C G q O E D F 0.36 a H B Ans

y FDB 0.785 x W 0.34 C F O Ox Oy B A F 0.11 Ans To find reactions at O
q=45 Find P ,O A FDB 0.785 x W F 0.11 0.34 C F O q Ox Oy a B To find reactions at O Ans

3/46 It is desired that a person be able to begin closing the van hatch from the open position shown with a 40-N vertical force P. As a desired exercise, determine the necessary force in each of the two gas-pressurized struts AB. The mass center of the 40-kg door is 37.5 mm directly below point A. Treat the problem as two-dimensional. Why direction: CB? 1125 mm A Ox Oy 550 mm 2F O W=mg 600 mm 175 mm P= 40N B Unknowns: F, Ox, Oy Equilibrium eqs:

ANS Find F 2F A 2F A O G G O W=(40)(9.81) W=mg B P = 40 B + Oy Ox y x
1125 mm 2F y A Oy 2F A 550 mm O G x G Ox O 600 mm W=(40)(9.81) 175 mm W=mg P = 40 B B + ANS

Assumption: no couple reactions develop at C,D
3/38 The hinge P fits loosely through the frame tube, and the frame tube has a slight clearance between the supports A and B. Determine the reactions on the frame tube at A and B, associated with the weight L of an 80-kg person. Also, calculate the changes in the horizontal reactions at C and D. Assumption: no couple reactions develop at C,D L= 80(9.81) ?

Find appropriate L to make
Bodies in equilibrium. : known q = : known x=?

3/52 The rubber-tired tractor shown has a mass of 13
3/52 The rubber-tired tractor shown has a mass of 13.5 Mg with the center of mass at G and is used for pusshing or pulling heavy loads. Determine the load P which the tractor can pull at a constant speed of 5 km/h up the 15-percent grade if the driving force exerted by the ground on each of its four wheel is 80% of the normal force under the wheel. Also find the total normal reaction N under the rear pair of wheels at B. Normal reaction at A and B are equal? If not, which one should be larger?

x y 600 1200 825 0.8 NA W P NA NB 0.8NB FBD W=mg=13.5*1000*9.81= N Ans

Recommended Problem 3/34 3/49 3/35 3/54 3/57

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