a Particle Forces the force. of action its sense is of application. Experimen demonstra forces ( P Resultant of Two Note: a) b) momentum)

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1 Chapter 2 : Statics of a Particle 2.2 Force on a Particle: Resultant of Two Forces Recall, force is a vector quantity whichh has magnitude and direction. The direction of the the force. force is defined byy the line of action and sense of The line of action is the infinite straight line acts. (Characterized by the angle it forms with along which the force some fixed axis.) The magnitude of the force is represented by the length of line, while its sense is indicated by the orientation n of the arrowhead. The spatial location of this force is generally epresented by its point of application. Experimen ntal evidence has demonstra ated that if two forces ( P & Q ) weree acting on a particle, they can be re eplaced R by a single force which is called the resultant force. Note: a) b) Scalars have magnitude only (e.g., time, mass, length, energy, etc.) Vectors have magnitude and direction (e.g., force, position, velocity, momentum) 1

2 2.3 Vectors Vectors are indicated by arrows placed above the letter, P scalars are without, P. and Note: In the text, vectors are in bold print. A vector used to epresent a force acting on well defined point of application, the particle. a given particle has a Such a vector is designated as a fixed or bound vector and cannot be moved without modifying the problem. Free vectors can be freely moved in space. 2.4 Addition of Vectors The addition of 2 vectors is commutative (Fig 2..7a&b) P + Q = Q + P Subtractio n of a vector is defined ass addition of a negative vector (See Fig 2.8a&b). P - Q = P + ( - Q ) Vectors which are coplanar reside in the same plane. 2

3 Vector addition is also associative. Therefore, the order in which vectors are added together is immaterial. P + Q + S = ( P + Q ) + S = P + ( Q + S ) Note vectors a and b are acting on a particle at point "P" a + b = c Useful trigonometric relationships Law of Sines sin = a sin = b sin c Law of Cosines a 2 = b 2 + c 2-2bc cos α b 2 = c 2 + a 2-2ca cos c 2 = a 2 + b 2-2ab cos γ 3

4 2.5 Resultant of Several Concurrent Forces If all the forces acting on a particle are coplanar and they all pass through the same point in space they are also said to be concurrent. R is the resultant of the 3 forces screen. P, Q, S residing in the plane of the 2.6 Resolution of a Force into Components The decomposition of a single force vector ( F ) into its components allows for easy manipulation. The use of vector decomposition is very convenient when trying to find a resultant vector. Although this may not seem helpful graphically y, it is analytically. Example: Find the resultant r and angle: r = a + b What to do?? 4

5 r = a + b r = ( a = x + b bx r = ( a ) + (aa y x + a r + b ) + ( b y ) x + y = r y ) x r + y b ) Let the magnitude of the vectors be: a = a =15 lb b = b = 20 lb Now compute the x-component: and So, a x = 15 cos 30 a x = lbs b x = 20 cos 70 b x = 6.84 lbss r x = a x + b x Repeat the processs for the y-component: a y = 15 sin 30 a y = 7.50 lbss and b y = 20 sin 70 b y = lbs r y = a y + b y 5

6 So the resultant vector is written as: r = r x î + r y ĵ r = ( ) î + ( ) ĵ r = (19.83 ) î + ( ) ĵ The magnitude of the vector is: r = r = (19.83) 2 + (26.30) 2 3 r = r = x 10 r = 32.9 N The resultant angle θ: tan = = arctan = -1 tan (1.326) = = The End --- 6

7 Sample problem 2.2, (part a) A barge is pulled by two tugboats. If the resultant force exerted by the boat is lbs directed along the x axis of the barge, determine the tension in each of the ropes if = 45. Find the tension T 1 and T2 in each rope. ==== The End ==== 7

8 HW #1: 2.1,7, and Unit Vectors and Force Components Vectors have properties of magnitude and orientation (or direction). The magnitude of a vector is a scalar (No directional information). Multiplicati ion of a vector by a scalar changes its magnitude only, not its orientation. A unit vector is one whose magnitude is unity. The use of unit vectors allows us to write vectors in two parts, one which represents the magnitude, and the otherr which epresentss the orientation n. NOTE: Therefore r = r êr indicates a unit vector in the r direction r and ê = r êr r = 1 ê êx = î = 1 ê êy = ĵ = 1 ê ez = kˆ = 1 Lets resolve the force F into itss 2 components 8

9 use the unit vectors î and ĵ to indicate the x and y direction F F x Fxiˆ F = F x and + F F x î + F Recall, we can determine magnitudes F x and F y by trigonometric relations F y y y ĵ F y ĵ F x = F cos θ and F y = F sin θ Example 2.7.1: A force of 800 N is exerted on the bolt shown. Find the corresponding componen nts. Write the Force vector in rectangula ar form. 9

10 Solution: ==== The End === == Example 2.7.2: A force vector; F = 700 lb î lb ĵ is applied to the Determine the magnitude and orientation of the resulting force vector. bolt. 10

11 ==== The End ==== 2.8 Addition of Forces by Summing x and y Components Resolving several forces acting at a point The resultant vector, R = P + Q + S 11

12 First find the x and y components of each vector P = P x î + P y ĵ ; Q = Q î + Q ĵ ; S = Sx î + Sy ĵ Therefore the resultant Force vector is, and R = (P x + Q R = R x x x î + R y +S x)î + (P y y ĵ + Q +S ) ĵ y y R x = ΣF x and R y = ΣF y ==== The End ==== 12

13 Example (Sample Prob 2.3): Determine the magnitude and direction of the resultant force acting on the bolt (Fig P23). Solution: 13

14 ==== The End ==== 14

15 Example (Problem 2.17): Determine the components of each of the forces shown as well as the resultant force vector, it's magnitude and orientation n. Solution: ==== The End === 15

16 Example (Problem 2.19): Determine the components of each of the forces shown as well as the resultant force vector, it's magnitude and orientation. Solution: ==== The End ==== 16

17 Example Problem 2.33: A hoist trolley iss subjected to the three forces shown. Knowing that P = 250 lb, determine e: a) the value of the angle α for which the resultant of the three forces is vertical, and b) the corresponding magnitude of the resultant. (The A force vector magnitude is 400 lbs and the B force vector is 2000 lbs). Solution: 17

18 ==== The End ==== HW #2: 2.16, 36, 51, and 52 18

19 2.9 Equilibrium of a Particle Recall Newton's Laws: When the resultant of all the forces acting on particle is in equilibrium. a particle is zero, the Therefore, a particle acted upon by two forces is in equilibrium when they have the same magnitude but actt in opposite directions. therefore; A x R = A + B R = = 0 R = (A x - B x = 0 - Bx) î + (Ay - and ) By ĵ = 0 A y - B y = 0 ΣF = 0 is also called static equilibrium Newton's First Law of Motion If the resulting force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion). 19

20 2.11 Free-Body Diagrams (Particles in Equilibrium) Space diagram - A sketch whichh shows the physical conditions of the problem. Free-body diagram - A separatee diagram showing the point of interest (or the particle) and all the forces acting on it. Force triangle - The closed triangle formed by three concurrent forces (acting at the same point) when drawn in a tip to tail fashion. 20

in order to center the automobile over its

The angle between the cable and the vertical

Solution: Free-Body Force Diagram Point A is

sines, sin = a sin sin = b c TAB sin 120 =

21 Example Sample Problem 2.4: In a ship-unloading operation, a 3500-lb automobile is supported by a cable. A rope is tied to the cable at A and pulled in order to center the automobile over its intended position. The angle between the cable and the vertical is 2, while the angle between the rope and the horizontal is 30. What is the tension in the rope? Solution: Free-Body Force Diagram Point A is chosen as the particle and a complete free body diagram is produced. Examine the force triangle using the law of sines, sin = a sin sin = b c TAB sin 120 = TAC sin 2 = 3500 lb sin lbs = lbs T A B = lbs sin120 T AB B = 3579 lb and TA C = lbs sin 2 T AC C = 144 lb ==== The End ==== 21

22 Problem 2.47: The 60 lb collar "A" slides on a frictionless vertical rod and is connected to the 65 lb counterweight C. Determinee the values of H x and "h" for which the system is in equilibrium. Solution: ==== The End ==== 22

23 2.12 Rectangul ar Force Compone ents in Space (GO TO File: DrDch02 Add_19 January2017) Now consider forces acting in 3 spatial dimension s x,y,z Let the plane OBAC contain force vector F F makes an with the y-axis y F = F cos y y & F h = F sin y F Fh may be still resolved into two components, one each in the x and z directions F x = F h cos and F z = F h sin Fx = F sin y cos and Fz = F sin y sin To find the resultant magnitude of 23

24 the vector F F F = F x î + F y ĵ + F Fz kˆ F = F = 2 2 = x y + 2 x F + F + F Instead of using the force F θ y and φ as previously suggested. Define the direction of by x, y and z ; these are called angles of the direction cosines. 24

25 START AGAIN They form the between the force vector and the coordinate axis. F = F(cos x î + cos y ĵ + cos z kˆ) So the force F may be expressed by a scalar F and a vector Therefore, = cos x î + cos y ĵ + cos z kˆ F = F The magnitude of is 1 and is in the same direction as F. Call a unit vector along the line of action of F. Therefore, x = cos x ; y = cos y ; z = cos z and iˆ ˆj kˆ x x + y + z = cos x + cos y + cos z =1 where cos x = Fx F ; cos y = Fy F ; cos z = Fz F 25

26 2.14 Addition of Concurrent Forces in Space The resultant force R = Rx î + R y ĵ + R z kˆ where R x = ΣF x ; R y = ΣF y ; R z = ΣF z and of course R = R = R 2 x + R 2 y + R 2 z We can write R = R where cos x = R R x & cos y = R R y & cos z = R R z HW #3: 2.64, 71, and 76 26

27 Sample Problem 2.7: The tension in the wire (F) is 2500 N. Find F x, F y, F z throughh the bolt at A, and θ x, θ y, and θ z defining the direction of the force. Solution: 27

28 ==== The End ==== 28

29 Problem 2.63: Find the resultant Force vector. Given: θ x = 70 & θ z = 130 & F y = +400 lb Solution: ==== The End ==== 29

30 Problem 2.67: The tension in cable AB is 285 lb. Find the F vector at " B". Solution: ==== The End == === 30

31 Problem 2.71: Find the Resultant Force vector F, it's components angles of the direction cosines. and Solution: 31

32 ==== The End ==== 32

33 2.15 Equilibrium of a Particle in Space According to Newton's 1st law; if the net force acting on a particle is zero the particle will not move and is said to be in equilibrium. Therefore, Rx = Fx 0 R R y z = Fy 0 (2.34) = F z 0 These are the necessary and sufficient conditions for equilibrium. HW #4: 2.88, 98, and

34 Problem 2.97 If the length of the cable is 1.5 m and its mass 300 kg. Find the tension in the cable. Solution: ==== The End == === 34

35 Problem 2.91: Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached to supports at D and E, respectively. Knowing that a 200 lb vertical load P is applied to ring A, determine the tension in each of the three cables. Solution: 35

36 ==== The End ==== 36

37 Problem 2.107: Determine the tensionn in each wire when a 180 lb. cylinder is suspended from point D. Solution: 37

38 ==== The End ==== 38

Chapter 2 Statics of Particles. Resultant of Two Forces 8/28/2014. The effects of forces on particles:

Chapter 2 Statics of Particles The effects of forces on particles: - replacing multiple forces acting on a particle with a single equivalent or resultant force, - relations between forces acting on a particle