ENGINEERING MECHANICS BAA1113

Transcription

1 ENGINEERING MECHANICS BAA1113 Chapter 3: Equilibrium of a Particle (Static) by Pn Rokiah Bt Othman Faculty of Civil Engineering & Earth Resources rokiah@ump.edu.my

2 Chapter Description Aims To explain the Equilibrium Equation To explain the Free Body Diagram To apply the Equations of Equilibrium to solve particle equilibrium problems in Coplanar Force System (2-D &3-D) Expected Outcomes Able to solve the problems of a particle or rigid body in the mechanics applications by using Equilibrium Equation References Russel C. Hibbeler. Engineering Mechanics: Statics & Dynamics, 14 th Edition

3 Chapter Outline 3.1 Equilibrium Equation 3.2 Free Body Diagram 3.3 Coplanar Force Systems (2-D &3-D) 3.4 Example

4 3.1 Equilibrium Equation What is Equilibrium? Equilibrium means the forces are balanced but not necessarily equal In physic, it means equal balance which the opposing forces or tendencies neutralize each other How to know the body is in Equilibrium? A body at rest or in uniform motion (velocity) is in equilibrium

5 Condition for the Equilibrium of a Particle How to know the body is in Equilibrium? Particle at equilibrium if - At rest - Moving at constant a constant velocity Newton s first law of motion F = 0 where F is the vector sum of all the forces acting on the particle T B = 4.90kN T D = 4.25kN + F x = 0; T B cos30 - T D = 0 + F y = 0; T B sin kn = 0

6 Condition for the Equilibrium of a Particle Newton s second law of motion F = ma When the force fulfill Newton's first law of motion, ma = 0 a = 0 therefore, the particle is moving in constant velocity or at rest Static Equilibrium is when the body at rest If the Dynamic Equilibrium, the body move and continue to move

7 Application of Equilibrium Equation Straps Construction-lifting the material Day life activity- Paddle the boat Support Oil Rig Playground- Seesaw

8 Application of Equilibrium Equation Measure the forces and length of cable AB and AC Measure the forces, direction and size of cable AB Measure the forces to make sure the rigging doesn t fail

9 3.2 Free Body Diagram (FBD) What is FBD? FBD is a sketch to show only the forces acting on selected body

10 3.2 Free Body Diagram (FBD) Best representation of all the unknown forces ( F) which acts on a body A sketch showing the particle free from the surroundings with all the forces acting on it Consider two common connections in this subject Spring Cables and Pulleys

11 Spring Linear elastic spring: change in length is directly proportional to the force acting on it spring constant or stiffness k: defines the elasticity of the spring Magnitude of force when spring is elongated or compressed F = ks where s is determined from the difference in spring s deformed length l and its undeformed length l o s = l - l s = l - l o o If s is positive, F pull onto the spring If s is negative, F push onto the spring

12 Cables and Pulley Cables (or cords) are assumed to have negligible weight and they cannot stretch A cable only support tension or pulling force Tension always acts in the direction of the cable Tension force in a continuouscable must have a constant magnitude for equilibrium For any angle θ, the cable is subjected to a constant tension T throughout its length With a frictionless pulley and cable T 1 = T 2.

13 3.3 Coplanar Systems 2-D A particle is subjected to coplanar forces in the x-y plane Resolve into i and j components for equilibrium F x = 0 F y = 0 Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal o (zero)

14 Scalar Notation Sense of direction = an algebraic sign that corresponds to the arrowhead direction of the component along each axis For unknown magnitude, assume arrowhead sense of the force Since magnitude of the force is always positive, if the scalar is negative, the force is acting in the opposite direction

15 3.3 Coplanar Systems 3-D When a particle is in equilibrium, the vector sum of all the forces acting on it must be zero ( F = 0 ) This equation can be written in terms of its x, y and z components.this form is written as follows ( F x ) i + ( F y ) j + ( F z ) k = 0 This vector equation will be satisfied only when F x = 0 F y = 0 F z = 0 These equations are the three scalar equations of equilibrium. They are valid for any point in equilibrium and allow you to solve for up to three unknowns.

16 Step to draw FBD Step 1: Sketch outline shape Step 2:Show all the forces that act on body and indicate the active (set the body in motion) or reactive forces (tend to resist the motion) Step 4: Apply EE and calculate the unknown forces( can be write in letters) Step 3: Labeled the known forces (magnitude and direction)

17 Select the correct FBD of Particle A 30 A lb A) A 100 lb F C) 30 A 100 lb B) D) F1 F2 30 A A 40 F2 F lb

18 FBD Using this FBD of Point C, the sum of forces in the x-direction ( F X ) is. Use a sign convention of lb C 50 F 2 A) F 2 sin = 0 F 1 B) F 2 cos = 0 C) F 2 sin 50 F 1 = 0 D) F 2 cos = 0

19 FBD Four forces act at point A and point A is in equilibrium. Select the correct force vector P. A) {-20 i + 10 j 10 k}lb B) {-10 i 20 j 10 k} lb C) {+ 20 i 10 j 10 k}lb D) None of the above. P F 1 = 20 lb x z F 3 = 10 lb F 2 = 10 lb A y In 3-D, when you don t know the direction or the magnitude of a force, how many unknowns do you have corresponding to that force? A) One B) Two C) Three D) Four

20 Example 3.1 This is an example of a 2-D or coplanar force system. If the whole assembly is in equilibrium, then particle A is also in equilibrium. Determine the tensions in the cables for a given weight of cylinder = 40kg Step 1: (Sketch outline shape)

21 Solution Example 3.1 Step 4: Apply EE and calculate the unknown forces( can be write in letters) FBD at A F D A A y 30 F B x Step 2:Show all the forces that act on body and indicate the active (set the body in motion) or reactive forces (tend to resist the motion) F C = 40 x 9.81 =392.4 N Step 3: Labeled the known forces (magnitude and direction)

22 Solution Example 3.1 Step 4: Apply EE and calculate the unknown forces( can be write in letters) FBD at A y Since particle A is in equilibrium, the net force at A is zero. So F B + F C + F D = 0 or F = 0 F D A A 30 F B x F C = N In general, for a particle in equilibrium, F = 0 or F x i + F y j = 0 = 0 i + 0 j (a vector equation) Or, written in a scalar form, F x = 0 and F y = 0 Two scalar equations of equilibrium (E-of-E) Used to solve for up to two unknowns

23 Solution Example 3.1 Write the scalar E-of-E: FBD at A y + F x = F B cos 30º F D = 0 + F y = F B sin 30º N = 0 F D A A 30 F B x Solving the second equation, F B = 785 N From the first equation, F D = 680 N F C = N

24 Example 3.2 This is an example of a 2-D or coplanar force system. If the whole assembly is in equilibrium, then particle E is also in equilibrium. Determine the tensions in the cables DE,EA and EB for a given weight of cylinder = 40kg Step 1: Draw point E Step 2 Step 3

25 Solution Example 3.2 FBD at point E y T EB = 40*9.81 N Step 2 T ED E 30 x T EA Applying the scalar E-of-E at E, Step 3 Step 4 + F x = T ED + (40*9.81) cos 30 = 0 + F y = (40*9.81) sin 30 T EA = 0 Solving the above equations, T ED = 340 N and T EA = 196 N

26 Example 3.3 This is an example of a 2-D or coplanar force system. If the whole assembly is in equilibrium, then particle C and D are in equilibrium. Determine the force in each cables for a given weight of lamp = 20kg How many unknown forces? Step 2: Draw point C to solve F CB & F CA Step 1: Draw point D to solve F CD & F DE

27 Solution Example 3.3 FBD at point D y F DE Step 1: Draw point D to solve F CD & F DE F CD D 30 x Applying the scalar E-of-E at D, + F y = F DE sin (9.81) = 0 + F x = F DE cos 30 F CD = 0 Solving the above equations, F DE = 392 N and F CD = 340 N W = 20 (9.81) N

28 Solution Example 3.3 F AC FBD at point C y Step 2: Draw point C to solve F CB & F CA C F CD =340 N x F BC 45 Applying the scalar E-of-E at C, + F x = 340 F BC sin 45 F AC (3/5) = 0 + F y = F AC (4/5) F BC cos 45 = 0 Solving the above equations, F BC = 275 N and F AC = 243 N

29 Example 3.4 This is an example of a 3-D or coplanar force system. If the whole assembly is in equilibrium, determine the tension developed in each cables T B How many unknown forces? Step 2: Write the unknown forces in cable T B,T C & T D in Cartesian vector Step 1: Draw point A to solve T B,T C & T D

30 Solution Example 3.4 T B = T B i T C = (T C cos 60 ) sin30 i + (T C cos 60 ) cos30 j + T C sin 60 k T C = T C (-0.25 i j k ) T B T D FBD at A T C T D = T D cos 120 i + T D cos 120 j +T D cos 45 k T D = T D ( 0.5 i 0.5 j k ) W = -300 k

31 Solution Example 3.4 Applying equilibrium equations: F R = 0 = T B i + T C ( 0.25 i j k ) + T D ( 0.5 i 0.5 j k ) 300 k T D FBD at A T C Equating the respective i, j, k components to zero, F x = T B 0.25 T C 0.5 T D = 0 (1) F y = T C 0.5 T D = 0 (2) F z = T C T D 300 = 0 (3) T B Using (2) and (3), T C = 203 lb, T D = 176 lb Substituting T C and T D into (1), T B = 139 lb

32 Example 3.5 This is an example of a 3-D or coplanar force system. If the whole assembly is in equilibrium, determine the tension developed in each cables How many unknown forces? Step 2: Write the unknown forces in cable T B,T C & T D in Cartesian vector Step 1: Draw point A to solve T B,T C & T D

33 Solution Example 3.5 FBD at A 2 m F D z F C 1 m 2 m A 30 y F B = F B (sin 30 i + cos 30 j) N x 600 N F B = {0.5 F B i F B j} N F C = F C i N F D = F D (r AD /r AD ) = F D { (1 i 2 j + 2 k) / ( ) ½ } N = { F D i F D j F D k } N

34 Solution Example 3.5 Now equate the respective i, j, and k components to zero F x = 0.5 F B F C F D = 0 F y = F B F D = 0 F z = F D 600 = 0 2 m F D 1 m 2 m x FBD at A z A 600 N F C 30 F B y Solving the three simultaneous equations yields F C = 646 N (since it is positive, it is as assumed, e.g., in tension) F D = 900 N F B = 693 N

35 Example 3.6 This is an example of a 3-D or coplanar force system. If the whole assembly is in equilibrium, and supported by two cables and strut AD. Given 400 lb crate, determine the magnitude of the tension developed in each cables and the force developed along strut ADs How many unknown forces? Step 2: Write the unknown forces in cable F B,F C & F D in Cartesian vector Step 1: Draw point A to solve F B,F C & F D

36 Solution Example 3.6 FBD of Point A z F C F B x F D W y W = weight of crate = k lb F B F C F D = F B (r AB /r AB ) = F B {( 4 i 12 j + 3 k) / (13)} lb = F C (r AC /r AC ) = F C {(2 i 6 j + 3 k) / (7)}lb = F D ( r AD /r AD ) = F D {(12 j + 5 k) / (13)}lb

37 Solution Example 3.6 The particle A is in equilibrium, hence FB + FC + FD + W = 0 Now equate the respective i, j, k components to zero (i.e., apply the three scalar equations of equilibrium). Fx = (4 / 13) FB + (2 / 7) FC = 0 (1) Fy = (12 / 13) FB (6 / 7) FC + (12 / 13) FD = 0 (2) Fz = (3 / 13) FB + (3 / 7) FC + (5 / 13) FD 400 = 0 (3) Solving the three simultaneous equations gives the forces FB = 274 lb FC = 295 lb FD = 547 lb

38 Conclusion of The Chapter 3 Conclusions - The Equilibrium and FBD have been identified - The Equilibrium Equation have been implemented to solve a particle problems in Coplanar Forces Systems

39 Credits to: Dr Nurul Nadhrah Bt Tukimat En Khalimi Johan bin Abd Hamid Roslina binti Omar