Engineering Mechanics : STATICS BDA10203 Lecture #08 Group of Lecturers Faculty of Mechanical and Manufacturing Engineering Universiti Tun Hussein Onn Malaysia

EQUILIBRIUM OF A PARTICLE IN 3-D Today’s Objectives: Students will be able to : a) Draw a free body diagram (FBD), and, b) Apply equations of equilibrium to solve a 3-D problem. Learning Topics: Application / relevance Equations of equilibrium

APPLICATIONS Whenever cables are used for hoisting loads, they must be selected so that they do not fail when they are placed at their points of attachment.

EQUATIONS OF 3-D EQUILIBRIUM (Section 3.4) Since particle A is in equilibrium, the net force at A is zero. So FB + FD + FC - W = 0 or  F = 0 In general, for a particle in equilibrium,  F = 0 or in cartesianvector equation Fx i + Fy j + Fz k = 0 i + 0 j + 0 k

EQUATIONS OF 3-D EQUILIBRIUM (Section 3.4) written in a scalar form, Fx = 0,  Fy = 0 and  Fz = 0 These are three scalar equations of equilibrium (EofE). They can be used to solve for up to three unknowns.

Given: Crate mass 100 kg. and geometry is as shown. EXAMPLE Given: Crate mass 100 kg. and geometry is as shown. Find: Forces in the cords AC, AD and the stretch of the spring. Plan: 1. Draw a FBD for Point A. 2. Apply EofE at Point A to solve for the unknowns (FB, FC & FD). 3. Evaluate s = FB/k. You may ask the students to give a plan You may explain why analyze at E, first, and C, later

EXAMPLE (continued) A FBD at A should look like the one to the left. Note the assumed directions for the three forces. From the FBD, one can write (the position vectors) FB = FB i FC = FC cos 1200 i + FC cos 1350 j + FC cos 600 k = -0.5FC i – 0.707 FC j + 0.5FC k FD = -0.333FD i + 0.667 FD j + 0.667FD k W = -981 k N You may ask students to give equations of equilibrium.students hand-outs should be without the equations.

EXAMPLE (continued) Equilibrium requires F = 0; FB + FC + FD + W = 0 FB i + -0.5FC i – 0.707 FC j + 0.5FC k + -0.333FD i + 0.667 FD j + 0.667FD k - 981 k = 0 The scalar EofE are: +   Fx = FB – 0.5FC -0.333FD = 0 (1) +   Fy = -0.707FC + 0.667FD = 0 (2) + z  Fz = 0.5FC + 0.667FD - 981 = 0 (3) You may ask students to give equations of equilibrium.students hand-outs should be without the equations.

EXAMPLE (continued) Solving these three simultaneous equations for the three unknowns yields: FB = 693.7 N FC = 813 N FD = 862 N The stretch of the spring is therefore F = ks; (or FB = ks) 693.7 = 1500 s s = 0.462 m You may ask students to give equations of equilibrium.students hand-outs should be without the equations.

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: Load 500N and geometry is as shown. Find: Forces in the cable AC, BC and CD. Plan: 1. Draw a FBD for Point C. 2. Apply EofE at Point C to solve for the unknowns (FCA, FCB & FCD).

GROUP PROBLEM SOLVING (continued) FCA = FCA UCA = FCA (rCA/rCA) rCA = {( XA – XC ) i + ( YA – YC ) j + ( ZA – ZC ) k }m = {(0.2 – 0 ) i + ( 0 – 0.6) j + ( 0 – 0) k }m = {0.2i - 0.6j + 0k }m FCA = FCA UCA = FCA (rCA/rCA) = FCA {0.32i – 0.95j + 0k}

GROUP PROBLEM SOLVING (continued) rCB = {( XB – XC ) i + ( YB – YC ) j + ( ZB – ZC ) k }m = {(0 – 0.2 ) i + ( 0 – 0.6) j + ( 0 – 0) k }m = {-0.2i - 0.6j + 0k }m FCB = FCB UCB = FCB (rCB/rCB) = FCB {-0.32i – 0.95j + 0k} rCD = {( XD – XC ) i + ( YD – YC ) j + ( ZD – ZC ) k }m = {(0 – 0) i + ( 1.2 – 0.6) j + ( 0.8 – 0) k }m = {0i + 0.6j + 0.8k }m FCD = FCD UCD = FCD (rCD/rCD) = FCD {0i + 0.6j + 0.8k}

GROUP PROBLEM SOLVING (continued) ΣFX = 0.32FCA – 0.32FCB = 0 ΣFy = -0.95FCA – 0.95FCB + 0.6FCD = 0 ΣFz = 0.8FCD - W = 0 After solving the equations above, the answers are as follow : FCD = 625 N FCB = 197.4 N FCA = 197.4 N

HOMEWORK TUTORIAL Q1 (3-45): The three cables are used to support the lamp of weight W. Determine the force developed in each cable for equilibrium. Units Used: kN = 1000N Given: W = 800N, b = 4m a = 4m, c = 2m

HOMEWORK TUTORIAL (continued) Q2 (3-47): Determine the stretch in each of the two springs required to hold the crate of mass 20kg in the equilibrium position shown. Each spring has an unstretched length of 2m and a stiffness of k = 300N/m. Given: a = 4m, b = 6m, c = 12m

HOMEWORK TUTORIAL (continued) Q3 (3-55): The ends of the three cables are attached to a ring at A and to the edge of the uniform plate of mass M. Determine the tension in each of the cables for equilibrium. Given: M = 150kg, e = 4m a = 2m, f = 6m b = 10m, g := 6m c = 12m, h := 6m d = 2m, i := 2m gravity = 9.81m/s²

HOMEWORK TUTORIAL (continued) Q4 (2-63): Determine the force in each cable needed to support the platform of weight W. Units Used: kN := 1000N Given: W = 17500N, d = 1m a = 1m, e = 1.5m b = 2m, f = 1.5m c = 2m, g = 5m