Chapter 2 Statics of Particles Addition of Forces

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+ 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 → → → 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

Rectangular Components of a Force Unit Vectors → → Unit vectors i , j

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

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

Addition of Forces by Summing Their components → P → S A → Q → → P+ Q + S = R

Addition of Forces by Summing Their components - Example Problem 2. 22 on page 33 Determine the resultant force applied on the bolt Force 7 k. N 9 k. N 5 k. N Magnitude x Component F 1 5 k. N 5* cos (40) = 5* sin (40) = 3. 83 k. N 3. 21 k. N F 2 7 k. N 7*cos(110)= -2. 39 k. N 7* sin (110)= 6. 57 k. N F 3 9 k. N 9*cos(160)= -8. 46 k. N 9* sin(160)= 3. 08 k. N -7. 01 k. N 12. 86 k. N R y Component

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 F 1 300 0 F 2 173. 2 0 -173. 2 F 3 200 -173. 2 F 4 400 -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 N 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.

Equilibrium of Particle – Example (continued) (1) N (2) N N 200*cos(240) = -100 200*sin(240) = -173. 2

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 C

Equilibrium of Particle – Example-2 For simplicity C

Equilibrium of Particle – Example-2 Since the system is in equilibrium The resultant force at C is zero. C

Equilibrium of Particle – Example-2 Since the system is in equilibrium The resultant force at C is zero. C

Equilibrium of Particle – Example-2 C (1) (2)

Equilibrium of Particle – Example-2 (1) C (2)

Forces in Space ( 3 D )

Forces in Space ( 3 D )

Forces in Space ( 3 D ) • The three angle 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 )

Forces in Space ( 3 D ) • The vector is the unit vector along the line of action of F

Forces in Space ( 3 D ) • The vector is the unit vector along the line of action of F

Forces in Space ( 3 D )

Forces in Space ( 3 D ) Direction of the force is defined by the location of two points,

Forces in Space ( 3 D )

Sample Problem 2. 7 The tension in the guy wire is 2500 N. Determine: a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles qx, qy, qz defining the direction of the force

Sample Problem 2. 7 • Determine the components of the force.

Sample Problem 2. 7 • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. or

Addition of Concurrent Forces in Space • The resultant R of two or more vectors in space

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

Problem 2. 103 on page 60