Engineering Mechanics : STATICS BDA10203 Lecture #04 Group of Lecturers Faculty of Mechanical and Manufacturing Engineering Universiti Tun Hussein Onn Malaysia

EQUILIBRIUM OF A PARTICLE IN 2-D Today’s Objectives: Students will be able to : a) Draw a free body diagram (FBD) b) Apply equations of equilibrium (EofE) to solve a 2-D problem.

Equilibrium of a Particle Particle at equilibrium if - At rest - Moving at 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

APPLICATIONS For a spool of given weight, what are the forces in cables AB and AC ?

APPLICATIONS (continued) For a given cable strength, what is the maximum weight that can be lifted ?

EQUILIBRIUM OF PARTICLE IN 2-D (Section 3.3) 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. To determine the tensions in the cables for a given weight of the engine, we need to learn how to draw a free body diagram and apply equations of equilibrium.

THE WHAT, WHY AND HOW OF A FREE BODY DIAGRAM (FBD) Free Body Diagrams are one of the most important things for you to know how to draw and use. What ? - It is a drawing that shows all external forces acting on the particle. Why ? - It helps you write the equations of equilibrium (EofE) used to solve for the unknowns (usually forces or angles).

2. Show all the forces that act on the particle. 1. Imagine the particle to be isolated or cut free from its surroundings. 2. Show all the forces that act on the particle. Active forces: They want to move the particle. Reactive forces: They tend to resist the motion. 3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables . Note : Engine mass = 250 Kg FBD at A

EQUATIONS OF 2-D EQUILIBRIUM (EofE) Since particle A is in equilibrium, the net force at A is zero. or  F = 0 So FAB + FAD + FAC = 0 In general, for a particle in equilibrium, Vector equation  F = 0 Fx i + Fy j = 0 Or, written in a scalar form, Fx = 0  Fy = 0 These are two scalar equations of equilibrium (EofE). They can be used to solve for up to two unknowns.

Solving the second equation gives: TB = 4.90 kN EXAMPLE 4.1 Note : Engine mass = 250 Kg FBD at A Write the scalar EofE: +   Fx = TB cos 30º – TD = 0 +  Fy = TB sin 30º – 2.452 kN = 0 Solving the second equation gives: TB = 4.90 kN From the first equation, we get: TD = 4.25 kN

EXAMPLE Given: Sack A weighs 20 N. and geometry is as shown. Find: Forces in the cables (EG,EC,CD) and weight of sack B (WB.) You may ask the students to give a plan You may explain why analyze at E, first, and C, later

PLAN 1. Draw a FBD for Point E. 2. Apply EofE at Point E to solve for the unknowns (TEG & TEC). 3. Repeat this process at C.

EXAMPLE (continued) The scalar EofE are: +   Fx = TEG sin 30º – TEC cos 45º = 0 +   Fy = TEG cos 30º – TEC sin 45º – 20 N = 0 You may ask students to give equations of equilibrium.students hand-outs should be without the equations. Solving these two simultaneous equations for the two unknowns yields: TEC = 38.6 N TEG = 54.6 N

EXAMPLE (continued) Now move on to ring C. The scalar EofE are:    Fx = 38.64 cos 45 – (4/5) TCD = 0    Fy = (3/5) TCD + 38.64 sin 45 – WB = 0 You may ask the students to give equations of equilibrium.Students hand-outs should be without the equation. You may also explain about the direction of 38.6 lb force (see fig. 3.7 d, page 88) Solving the first equation and then the second yields TCD = 34.2 N and WB = 47.8 N .

SPRINGS, CABLES, AND PULLEYS Spring Force = spring constant * deformation, or F = k * S With a frictionless pulley, T1 = T2.

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 = kx

Spring where x is determined from the difference in spring’s deformed length l and its undeformed length lo x = l - lo - If x is positive, F “pull” onto the spring - If x is negative, F “push”

Spring Example x = l - lo Given lo = 0.4m and k = 500N/m To stretch it until l = 0.6m, F = kx =(500N/m)(0.6m – 0.4m) = 100N is needed To compress it until l = 0.2m, =(500N/m)(0.2m – 0.4m) = -100N is needed x = l - lo

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 continuous cable must have a constant magnitude for equilibrium

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: The car is towed at constant speed by the 600 N force and the angle  is 25°. Find: The forces in the ropes AB and AC.

GROUP PROBLEM SOLVING (continued) 30° 25° 600 N FAB FAC A FBD at point A Applying the scalar EofE at A, we get; +  Fx = FAC cos 30° – FAB cos 25° = 0 +  Fy = -FAC sin 30° – FAB sin 25° + 600 = 0 Solving the above equations, we get; FAB = 634 N FAC = 664 N

ATTENTION QUIZ 2. Using this FBD of Point C, the sum of forces in the x-direction ( FX) is ___ . Use a sign convention of +  . A) F2 sin 50° – 20 = 0 B) F2 cos 50° – 20 = 0 C) F2 sin 50° – F1 = 0 D) F2 cos 50° + 20 = 0 F2 20 N 50° C F1 Answers: 2. B

HOMEWORK TUTORIAL Q1 Determine the magnitudes of F1 and F2 so that the particle is in equilibrium.

HOMEWORK TUTORIAL (continued) Q2 (2.46) : Two cables are tied together at C and are loaded as shown. Knowing that α = 30°, determine the tension (a) in cable AC, (b) in cable BC.

HOMEWORK TUTORIAL (continued) Q4 (2.67) : A 280-kg crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (Hint: The tension in the rope is the same on each side of a simple pulley.) T T T T T (a) (b) (c) (d) (e)

HOMEWORK TUTORIAL (continued) Q5 (3-13) : Determine the stretch in each spring for equilibrium of the block of mass M. The springs are shown in the equilibrium position. Given: M = 2kg a = 3m b = 3m c = 4m kAB=30N/m kAC = 20N/m kAD = 40N/m g = 9.81ms²

HOMEWORK TUTORIAL (continued) Q6 (3-17) : Determine the force in each cable and the force F needed to hold the lamp of mass M in the position shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the equilibrium at C. Given: M := 4kg θ1 := 30° θ2 := 60° θ3 := 30°