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EGR 231 Engineering Statics: Fall 2011 Lecture 07: Equilibrium in 2-D Today: Homework Questions? Newton’s 2nd Law 2-D equilibrium problems. 500N Homework Problem Assignment 07: Problem 2.44 Two cables are tied together at C and are loaded as shown. Determine the tension a) in cable AC, b) in cable BC. y FB Problem 2:51 A welded connection is in equilibrium under the action of the four forces shown. Knowing that F A=8 kN and FB = 16 kN, determine the magnitudes of the other two FC forces. x 5 3 4 FA FD Problem 2.67: A 600 lb crate is supported by several rope and pulley arrangements as shown. Determine for each arrangement the T T tension in the rope. T T T (a (b (c (d (e ) ) ) ) ) T T T T T Following Today’s Lecture you should be able to: 1) Write Newton’s 2nd law for the case of static equilibrium of particles. 2) Set up and find the forces needed to give static equilibrium of a particle. Particle Equilibrium in 2-Dimensions: A particle is in equilibrium if it is at rest, or if it has a constant velocity. When in equilibrium, the sum of all forces acting on a body is zero. F 0 Free Body Diagram: A diagram of a body which shows all forces that at upon it and replaces all connections to the outside with other forces. Step 1: Isolate the body from the surroundings. Step 2: Show all forces. Include pure forces and all reactive forces due to cut or removed constraints. Step 3: Label know forces with values. Label unknown forces with variables. The equation F 0 is a vector equation. In 2-dimensions this can be broken down into two scalar equations... one in the x- and one in the y- directions. F 0 F1 F2 F3 0 ˆ ˆ ˆ F1x i F1 y ˆ F2 x i F2 y ˆ F3x i F3 y ˆ 0 j j j ˆ ( F1x F2 x F3x )i ( F1 y F2 y F3 y ) ˆ 0 j F x i Fy ˆ 0 ˆ j or breaking it down into two scalar equations: x-dir: F x 0 and y-dir: F y 0 y Example Problem 1: F Find the force F required FA=150 lbf θ to provide static equilibrium x of the green particle. O 30o FB=200 lb -------------------------------------------------------------------------------- Given: FA 150 ˆ + 0ˆ lb i j F 200sin 30 ˆ 200cos30 ˆ lb B i j 100 ˆ 173.2ˆ lb i j F F cosθ ˆ F sinθˆ i j Find the force, F that makes the system in equilibrium: F 0 FA FB F 0 (150 ˆ + 0ˆ) (100 ˆ 173.2 ˆ) ( F cos θ ˆ F sin θˆ ) 0 i j i j i j (150 100+ F cos θ) ˆ ( 0 173.2 F sin θ)ˆ 0 i j i: 0 150 100+ F cos θ j: 173.2 F sin θ 0 F cosθ=50 F sin θ 173.2 To solve divide the 2nd equation by the 1st. F sin θ 173.2 tan θ =3.464 θ 73.9o F cos θ 50 Therefore 50 50 F= 180.3 lb so F =180.3 lb at 73.9 o cos θ cos 73.9 Example Problem 2 B C Find the maximum load value 5 3 which does not exceed a cable 60o A 4 tension of 780 lbf in either cable. ------------------------------------------------------------------ Given: W 4 3 y Draw FBD FC FC ˆ + FC ˆ i j 5 5 FB 5 FB FB cos60 ˆ FB sin 60 ˆ FC 3 i j 4 W 0 ˆ Wˆ i j θ 60o x For Equilibrium: A F 0 FC FB W 0 W 4 3 ( FC ˆ + FC ˆ ) ( FB cos 60 ˆ FB sin 60 ˆ ) ( W ˆ ) 0 i j i j j 5 5 4 3 i: 0 FC FB cos 60 j: 0 FC FB sin 60 W 5 5 3 FC 0.625 FB W FC FB sin 60 5 At this point we can see that FC < FB, so let FB = 780 lbf and then calculate the other forces. FB 780 lb FC 0.625 FB 0.625(780) 487.5 lb f 3 3 W FC FB sin 60 (487.5) 780sin 60 968lb f 5 5 This clearly shows the rope holding the weight exceeds 780 lb, so redo the calculations letting W = 780 lb. 3 3 780 FC FB sin 60 (0.625FB ) FB sin 60 1.241FB 5 5 780 FB 628.5 lb and FC 0.625(628.5) 392.8 lb 1.241 Example 3 The 40 N collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The spring is unstretched when h = 300 mm. Knowing that the constant of the spring is 560 N/m, determine the value of h for which the system is in equilibrium. ------------------------------------------------------------------------------- B h k=560 N/m A 300 mm C Example 3 Solution: The 40 N collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The spring is unstretched when h = 300 mm. Knowing that the constant of the spring is 560 N/m, determine the value of h for which the system is in equilibrium. ------------------------------------------------------------------------------- B Geometry FBD Fs 300mm h θ k=560 N/m 300 L1 mm N A h A L2 300 mm C W = 40 N d θ Vector Forces: FS Fs sinθ ˆ + Fs cosθ ˆ i j N N ˆ 0ˆ i j W 0 ˆ 40ˆ N i j Spring Equation: FS k L k ( L2 L1 ) 560( L2 L1 ) Equilibrium: F 0 FS N W 0 Fs sin θ ˆ + Fs cosθ ˆ N ˆ 40ˆ 0 i j i j i: Fs sin θ N 0 j: Fs cos θ 40 0 N Fs sin θ Fs cos θ 40 Geometry relationship: 300mm L1 0.3002 0.3002 0.4243m 300 L1 L2 0.300 (0.300 d ) 2 2 mm h 0.3 d cos θ L2 L2 d θ 0.3 sin θ L2 So now start combining the equations: Fs cos θ 40 FS 562( L2 L1 ) become 0.3 d Fs 40 FS 562( L2 0.4243) L2 and combining these two give 40 ( L 0.4243)(0.3 d ) 0.4243 2 1 (0.3 d ) 562 L2 L1 Since L2 and d are related by L2 0.3002 (0.300 d )2 we finally get a single equation with a single variable, but it's not fun to solve being nonlinear in form. 0.4243 0.071174 1 (0.3 d ) 0.3002 (0.300 d )2 Rewrite as a function equation to zero. 0.4243 f (d ) 0 1 (0.3 d ) 0.071174 0.3002 (0.300 d )2 This can be solved by plotting the function and locating the root. 0.4243 f (d ) 0 1 (0.3 d ) 0.071174 0.3002 (0.300 d )2 Matlab code used to generate graph: L1=0.4243 k=562 s=0.3 W=40 d=linspace(0.11,0.12) x=(1-(L1./sqrt(s.^2+(s+d).^2))).*(s+d)-W/k h1=plot(d,x,'b') set(h1,'linewidth',2.0) xlabel('d') ylabel('residual') grid on Therefore: d = 0.115 or h = d + 0.3 = 0.415 m ------------------------------------------------------------------------------------------------------ An alternative method is to use Mathematica (available at LANDesk Management download site) to solve the equation. Using the Mathematica Solve command: Example 4: y FB An irregularly shaped machine component is held in the position shown by three clamps. Knowing that Fa=940 N, determine the magnitudes of three forces Fb and Fc exerted by the other two clamps FA 50o x 70o -------------------------------------------------------------------------------- Given: FA 940 ˆ + 0ˆ N FC i j F F cos50 ˆ F sin50 ˆ B B i B j 0.6428FB ˆ 0.7660FB ˆ i j FC FC cos70 ˆ FC sin 70 ˆ i j 0.3420F ˆ 0.9397F ˆ i C j C Find the forces, FB and FC that give the system equilibrium: F 0 FA FB FC 0 ( 940 ˆ + 0ˆ ) ( 0.6428FB ˆ 0.7660FB ˆ ) ( 0.3420FC ˆ 0.9397 FC ˆ ) 0 i j i j i j ( 940 0.6428FB 0.3420FC ) ˆ +( 0 0.7660FB 0.9397FC )ˆ 0 i j i: 0 940 0.6428FB 0.3420 FC j: 0 0.7660 FB 0.9397 FC 0 940 0.6428FB 0.7660 FB FC FC 0.3420 0.9397 FC 2749 1.8795FB FC 0.8152 FB To solve set the two expressions for FC equal: 2749 2749 1.8795FB 0.8152 FB FB 1020 N 0.8152 1.8795 then FC 0.8152 FB 0.8152(1020) 832 N Workout Problem: Two cables are tied together at C and are loaded as shown. Determine the tension a) in cable AC, b) in cable BC.