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Chapter 2 Statics of Particles Addition of Forces Parallelogram Rule: The addition of two forces P and Q : Draw two parallel lines of vectors to form a parallelogram Draw the diagonal vector represent the resultant force → P A → Q → → → P+ Q = R Addition of Forces Triangle Rule: Alternative method to Parallelogram Rule Arrange two vectors in tip-to-tail fashion Draw the vector from starting point to the tip of second vector → P → → → A → Q P+ Q = R Addition of Forces Triangle Rule: The order of the vectors does not matter → → → P → P+ Q = R A → Q Addition of two forces is commutative → → → → P + Q = Q+ P Addition of Forces More than two forces: Arrange the given vectors in tip-to-tail fashion Connect the tail of first vector to the tip of last vector → → → → P+ Q + S = R → S → P A → Q Addition of Forces More than two forces: The order of the vectors does not matter → → Q P → P → → → → → S P+ Q + S = R → S A A → Q → → → → → → + = S+ P Q Q+ S + P → → → = P+ Q + S Rectangular Components of a Force → → Fx , Fy + → → F = Fx → Rectangular components of F θ : Angle between x-axis and F measured from positive side of x-axis → Fy Rectangular Components of a Force Unit Vectors → → Unit vectors i , j Fx Fx i F y Fy j F Fx i Fy j Rectangular Components of a Force Unit Vectors → → Unit vectors i , j Fx Fx i F y Fy j F Fx i Fy j F F F 2 x 2 y Fx F cos Fy F sin Rectangular Components of a Force Example A Force of 800 N is applied on a bolt A. Determine the horizontal and vertical components of the force Fx F cos 800cos(1450 ) 665 N Fx (665 ) i Fy F sin 800 sin(1450 ) 459 N Fy (459 ) j F (655) i (459) j Rectangular Components of a Force Example A man pulls with a force of 300 N on a rope attached to a building. Determine the horizontal and vertical components of the force applied by the rope at point A AB 8 2 6 2 10 8 8 cos 0.8 36.860 AB 10 360 36.86 323.140 Fx F cos 300 cos(323.140 ) 240 N Fy (459 ) j Fy F sin 300sin(323.140 ) 180 N Addition of Forces by Summing Their components → → P S A → Q → → → → P+ Q + S = R Px i Py j Qx i Qy j S x i S y j Rx i Ry j R x ( Px Q x S x ) R y ( Py Q y S y ) R x Fx R y Fy Addition of Forces by Summing Their components - Example Problem 2.22 on page 33 Determine the resultant force applied on the bolt Force 7 kN 5 kN 9 kN Magnitude x Component F1 5 kN 5* cos (40) = 5* sin (40) = 3.83 kN 3.21 kN F2 7 kN 7*cos(110)= -2.39 kN 7* sin (110)= 6.57 kN F3 9 kN 9*cos(160)= -8.46 kN 9* sin(160)= 3.08 kN -7.01 kN 12.86 kN R y Component R (7.01 kN ) i (12.86 kN ) j Equilibrium of Particle When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium Equilibrium of Particle - Example Determine the resultant force applied on point A N N N Force Magnitude x Component y Component F1 300 300 0 F2 173.2 0 -173.2 F3 200 -100 -173.2 F4 400 -200 346.4 0 0 N 200*cos(240) = -100 200*sin(240) = -173.2 R Equilibrium of Particle – Newton’s First Law If the resultant force acting on a particle is zero, • the particle remains at rest (if originally at rest) • the particle moves with a constant speed in a straight line ( if originally in motion) Equilibrium State Equilibrium of Particle - Example Load with mass of 75 kg Determine the tensions in each ropes of AB and AC W g m ( 9.81 m / s 2 ) ( 75 kg ) 736 N N Rx Fx TAB x TAC x 0 N N 200*cos(240) = -100 200*sin(240) = -173.2 Since the load is in equilibrium state, The resultant force at A is zero. TAB cos(1300 ) TAC cos(300 ) 0 Ry Fy TAB y TAC y 0 TAB sin(1300 ) TAC sin(300 ) 736 0 Equilibrium of Particle – Example (continued) T AB cos(1300 ) T AC cos(300 ) 0 (0.643) T AB (0.866) T AC 0 N TAB sin(1300 ) TAC sin(300 ) 736 0 (0.766) TAB (0.500) TAC 736 0 (2) N N TAB 647 N TAC 480 N 200*cos(240) = -100 200*sin(240) = -173.2 (1) Equilibrium of Particle – Example-2 Two cables are tied together at C and they are loaded as shown Determine the tensions in cable AC and BC Problem 2.44 / page 41 Equilibrium of Particle – Example-2 Draw “Free Body” Diagram TBC T AC C W 3 kN Equilibrium of Particle – Example-2 For simplicity F1 TBC T AC F2 C W 3 kN F3 Equilibrium of Particle – Example-2 Since the system is in equilibrium The resultant force at C is zero. F2 F1 1 2 C F3 3 kN Ry Fy F1y F2 y F3 y 0 F1 sin(1 ) F2 sin( 2 ) F3 sin(3 ) 0 Re memberthe definitionof F1 sin(180 ) F2 sin( 2 ) F3 sin(2700 ) 0 Equilibrium of Particle – Example-2 Since the system is in equilibrium 2 The resultant force at C is zero. F2 F1 1 2 C F3 3 kN 500 0 43.6 525 300 0 Arc tan 36.86 400 2 Arc tan 1 180 36.86 143.130 F1 sin(143.130 ) F2 sin(43.60 ) F3 sin(2700 ) 0 Equilibrium of Particle – Example-2 F2 F1 C 2 1 F1 sin(143.130 ) F2 sin(43.60 ) F3 sin(2700 ) 0 (0.6) F1 (0.69) F2 3000 0 (1) F3 3 kN Rx Fx F1x F2 x F3x 0 F1 cos(1 ) F2 cos( 2 ) F3 cos(3 ) 0 F1 cos(143.130 ) F2 cos(43.60 ) F3 cos(2700 ) 0 (0.8) F1 (0.724) F2 0 (2) Equilibrium of Particle – Example-2 F2 F1 C 2 1 (0.6) F1 (0.69) F2 3000 0 (0.8) F1 (0.724) F2 0 F3 3 kN F1 2202 N F2 2433N (1) (2) Forces in Space ( 3 D ) Fx F cos( x ) Fy F cos( y ) Forces in Space ( 3 D ) Fz F cos( z ) Forces in Space ( 3 D ) Fx F cos( x ) Fy F cos( y ) Fz F cos( z ) •The three angle x , y , z define the direction of the force F •They are measured from the positive side of the axis to the force F •They are always between 0 and 180º Forces in Space ( 3 D ) F Fx i Fy j Fz k Forces in Space ( 3 D ) •The vector is the unit vector along the line of action of F F vector F its m agnitude F cos x i F cos y j F cos z k F F (cos x i cos y j cos z k ) F cos x i cos y j cos z k Forces in Space ( 3 D ) F Fx i Fy j Fz k F cos x i F cos y j F cos z k F cos x i cos y j cos z k F cos x i cos y j cos z k •The vector is the unit vector along the line of action of F Forces in Space ( 3 D ) Fx F cos( x ) Fy F cos( y ) Fz F cos( z ) When thecomponentsFx , Fy and Fz of a forceF are given F Fx2 Fy2 Fz2 Fx cos x F cos y Fy F Fz cos z F Forces in Space ( 3 D ) Direction of the force is defined by the location of two points, M x1 , y1 , z1 and N x2 , y2 , z 2 Forces in Space ( 3 D ) d vectorjoining M and N d xi d y j d z k where d x x 2 x1 d y y 2 y1 d d x2 d y2 d z2 d z z 2 z1 F F 1 FF d xi d y j d z k d Fd x Fd y Fd z F i j k d d d Fd y Fd x Fd z Fx Fy Fz d d d d d 1 d xi d y j d z k d Sample Problem 2.7 A (40, 0, 30) B (0, 80, 0) d x x 2 x1 0 40 40m d d y y 2 y1 80 0 80m d z z 2 z1 0 (30) 30m d vectorjoining A and B d -( 40m) i ( 80m) j ( 30m) k The tension in the guy wire is 2500 N. d d x2 d y2 d z2 94.3m Determine: d a) components Fx, Fy, Fz of the force d acting on the bolt at A, 1 40i 80 j 30k b) the angles x, y, z defining the 94.3 direction of the force 0.424i 0.848 j 0.318k Sample Problem 2.7 0.424i 0.848j 0.318k • Determine the components of the force. F F 2500 N 0.424 i 0.848 j 0.318k 1060 N i 2120 N j 795 N k Sample Problem 2.7 • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. cos x i cos y j cos z k 0.424 i 0.848 j 0.318k x 115.1 y 32.0 z 71.5 or cos x Fx 1060N F 2500N cos y Fy F 2120 2500 cos z Fz 795 F 2500 Addition of Concurrent Forces in Space • The resultant R of two or more vectors in space R Forces Rx F x ' s Ry F y ' s Rz F z ' s R R x2 R y2 R z2 R cos x x R cos y Ry R cos z Rz R Sample Problem 2.8 m A concrete wall is temporarily held by the cables shown. m The tension is 840 N in cable AB and 1200 N in cable AC. m m Determine the magnitude and direction of the resultant vector on stake A Sample Problem 2.8 Equilibrium of a Particle in Space When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium F x 0 F y 0 F z 0 Problem 2.103 on page 60