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# Vector lesson and problems

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### Vector lesson and problems

1. 1. ENGT 215 - StaticsChapter 2 – Resultant of Coplanar Force Systems
2. 2. Chapter Goals Force System – any set of forces treated as a group. Equivalent Force Systems – any two systems of forces which have the same mechanical effect on a body. Resultant – a single force that is equivalent to a given force system (same mechanical affect on the body). Force System Resultant  The goal of this chapter is to learn about equivalent force systems and how to calculate the resultants for planar force systems.
3. 3. Vector RepresentationAnatomy of a Force Vector: Represented by arrow AB showing the line of action. Length of AB represents force magnitude at some convenient scale. Direction for line of action is indicated by angle θ counter-clockwise from positive “x” reference axis (Standard Position). Arrowhead indicates sense of force. Reference coordinate system (x-y axes) is established at the point of application A. Tip B F Tail A Point of application
4. 4. Vector Representation Equal Vectors - Two vectors are equal if they have the same magnitude AND the same direction as in (a). The line of action may be different as shown. NegativeVectors - Two vectors are negatives of each other if they have the same magnitude and opposite directions as in (b).
5. 5. Resultant of Concurrent ForcesAdding Vectors – Finding Resultants: The resultant of multiple vectors is the vector sum of those vectors. Vectors cannot be added algebraically, they must be added geometrically! Two Parallelogram Rulefor  common methods adding vectors (finding the resultant); Parallelogram - a planar four sided figure with parallel opposite sides.  Triangle Rule
6. 6. Resultant of Concurrent ForcesParallelogram Rule - Method for Adding Vectors : Recreate the two vectors to be added such that their tails are coincident. The direction and magnitude of each vector should not change! Create a parallelogram by drawing two lines, one from the tip of each vector and parallel to the other vector. The intersection of these lines is the tip of the resultant vector R = v + w. Create the resultant by drawing a new vector R.  Tail of R should correspond with tail of the other vectors v and w.  Tip of R will be where construction lines intersect. *Note – magnitude of R is not the sum of the magnitudes of v and w.
7. 7. Resultant of Concurrent ForcesParallelogram Rule: Propertiesof parallelograms useful for calculating force resultants:  Sum of the interior angles (2B + 2C) is 360 o.  Opposite sides are equal in length.  ∠A + ∠B = 180º  ∠A = ∠C C B A B C
8. 8. Resultant of Concurrent ForcesTriangle Rule - Method for Adding Vectors: Recreate the two vectors such that the tail of second vector w is coincident with the tip of the first vector v. The direction and magnitude of each vector should not change! Createthe resultant by drawing a new vector R.  The tail of R should correspond with tail of first vector v.  The tip of R should correspond with tip of second vector w. Force Triangle – triangle formed by forces.
9. 9. Resultant of Concurrent ForcesPolygon Rule: (Polygon – any closed figure with straight sides) An extension of the Triangle Rule. Sum of 3 or more coplanar vectors can be accomplished by adding 2 vectors successively, forming a polygon. Example: R = a + b + c  R = (a + b) + c Force Polygon – polygon formed by forces. b R = (a + b) + c a a +b c R
10. 10. Resultant of Concurrent ForcesPolygon Rule: A more general form of Polygon Rule for adding vectors:  Recreate all summed vectors tip-to-tail.  Create the resultant vector R.  The tail of R should correspond with the tail of the first vector v1.  The tip of R should correspond with the tip of the last vector v6.
11. 11. Resultant of Concurrent Forces VectorSums are Commutative:  Get the same result using (v + w) or (w + v).  Applies to all methods for adding vectors. R=w+v R v w
12. 12. Resultant of Concurrent ForcesSubtracting Vectors:A vector can be subtracted by adding its negative. Remember, the negative of a vector has the same magnitude but the opposite direction. R = v + (-w) Subtracting Vectors
13. 13. Resultant of Concurrent ForcesResultant: a system of concurrent coplanar forces acting on a rigid body may be replaced by a single Resultant force which equals the vector sum of the given forces.The vector sums can be determined by two methods: Graphical Method –  Use linear scale and protractor to lay out vectors as previously described.  Measure length and angle of resultant.  Too Inaccurate! Don’t Use! Trigonometric Method –  Lay out vectors as described but not necessarily to scale.  Label the known angles and lengths of sides (magnitudes).  Compute the resultant using trigonometry.  Law of Sines  Law of Cosines  This is the method to be used in this class!
14. 14. Example 1: Determine the resultant of the two forces F 1 and F2 acting on the hook. F1 = 54N 60o F2 = 60N
15. 15. Example 1: Solution  Determine the resultant of the two forces F1 and F1 = 54N F2 acting on the hook. 60o1. Draw the vector diagram. F2 = 60N2. Use the Law of Cosines to find magnitude R.3. Use the Law of Sines to find direction of R.R 2 = F12 + F22 − 2 F1 F2 cos φ Law of CosinesR 2 = 60 2 + 54 2 − 2 ⋅ 60 ⋅ 54 ⋅ cos(120)R = 98.77 ≈ 98.8 N R = 98.8 N ∠ 28.3°sin β sin 120 =  Law of Sines R F1 = 54N F1 R 120o F1 β ϕ 60osin β = sin 120 R F2 = 60Nβ = 28.259o
16. 16. Example 2: Given: when θ = 30º, the resultant of F1 and F2 is 1 kN in the vertical downward direction. Find: the value of F1 and F2. Ropes support an I-beam of weight 1 kN. Thus resultant is 1 kN vertically downward.
17. 17. Example 2: Solution  Given: when θ = 30º, the resultant of F1 and F2 is 1 kN in the vertical downward direction.  Find: the value of F1 and F2. Solution: 1. Draw the vector diagram 2. Use the sine law to find F1 and F2 F1 1000 o = o ∠ ⇒ F1 = 653 N 290ºSin30 Sin130 F2 1000 o = o ⇒ F2 = 446 N 240º ∠Sin 20 Sin130
18. 18. Exercise 1: Determine the magnitude and direction of the resultant of the two forces acting on the eye hook. Hint: The law of cosines is handy.
19. 19. Exercise 1: Solution Determine the magnitude and direction of the resultant of the two forces acting on the eye hook.Magnitude by Law of Cosines:R2 = F12 + F22 – 2 F1F2cos 30 YR2 = 852 + 1502 – 2 (85)(150)cos 30 F2 = 150 NR2 = 7641.4 N2R= 87.4 N R θ 30º X Angle by Law of Cosines: F1 = 85 N F22 = F12 + R2 – 2 F1Rcosθ cosθ = (F12 + R2 – F22 ) / (2F1R) *Note: to determine an angle > 90º such cosθ = (852 + 87.42 – 1502 ) / (2*85*87.4) as θ, use the law of cosθ = -0.5139 Cosines instead of θ = 120.9º R = 87.4 N ∠ 120.9° the law of Sines.
20. 20. Exercise 2: Determine the magnitude and direction of the resultant of the two forces acting on the eye hook.
21. 21. Exercise 2: Solution Determine the magnitude and direction of the resultant of the two forces acting on the eye hook.Magnitude by Law of Cosines: YR2 = F12 + F22 – 2 F1F2cos 35 F2 = 65 NR2 = 802 + 652 – 2 (80)(65)cos 35 RR2 = 2105.82 N2 25°R= 45.8892 = 45.9 N α 10° F 1 = 80 N θ 10° X Angle by Law of Sines: 30º (sin α / 65 N) = (sin 35° / 45.8892 N) sin α = (65)(sin 35°) / 45.8892 N *Note: to determine α = 54.3355° an angle > 90º such as θ, use the law of θ = α + 10º Cosines instead of θ = 64.3º R = 45.9 N ∠ 64.3° the law of Sines.
22. 22. End Lesson 5
23. 23. Rectangular Components Force Component: A resultant is a single force that is equivalent to the sum of a group of forces. In other words it can replace all those forces and give the same effect. The individual forces that the resultant replaces are called the components of the resultant. F1 F Example: F1 and F2 are components of resultant F F2 Rectangular Components: 2 mutually perpendicular components.  Typically selected along horizontal x and vertical y axes.  Think of F and F as the projection of F onto the X and Y axes. x y y Example: Fx and Fy are Fy F rectangular components of F θ x 0 Fx + Fy = F Fx
24. 24. Rectangular ComponentsRectangular Components: Rectangular components are useful because they isolate the effect of the resultant force in each direction. The effect of the force in the X direction acts independently of the force effect in the Y direction (and vice versa). Using Rectangular Components, the effect of the resultant force in each direction can be treated independently. Example: Non-rectangular components F1 F F1 and F2 both have a force effect in the X direction. F2 y Example: Rectangular components F Only Fy has an effect in Y direction and Fy θ x only Fx has an effect in the X direction. 0 Fx They can be treated independently.
25. 25. Rectangular ComponentsRectangular Components Magnitude of rectangular components: If the magnitude and direction of F is known, the magnitudes of the rectangular components can be calculated from right triangle trig. Magnitude Fx = F cos θ Fx is called the x-component of F. Fy = F sin θ Fy is called the y-component of F. y y Fy F F Fy θ x θ x 0 0 Fx Fx
26. 26. Rectangular Components yRectangular Components Magnitude of Rectangular Components: F Fy What if F is directed as shown? θ x Fx 0 Assume Standard Position:  Use Reference Angle (α): y y Fy F F Fy θ α θ x x 0 0 Fx FxAngle θ measured from +x axis. Use α (positive acute angle from +xEquations yield direction sign of or –x axis). component. Direction sign found by inspection.Example: if θ = 150° Example: if θ = 150°, α = 30°Fx = F cos 150 = -0.866 F Fx = F cos 30 = 0.866 F  -0.866 FFy = F sin 150 = 0.500 F Fy = F sin 30 = 0.500 F
27. 27. Rectangular Components – Reference AngleFor any angle θ in standard position, the Reference Angle α(pronounced “alpha”) is defined as the acute positive anglebetween the vector and the x-axis.•Reference Angle α is acute (0º < α < 90º).•Reference Angle α is always positive. II To Calculate Reference Angle α II 0º < θ < 90º, then α = θII 90º < θ < 180º, then α = 180º - θIII 180º < θ < 270º, then α = θ - 180ºIV 270º < θ < 360º, then α = 360º - θ III IV
28. 28. Example 3: Resolve the 60 N force on the ring into its horizontal (x) and vertical (y) components. 60° F = 60N
29. 29. y Example 3: Solution Fx Resolve x the 60 N force on the ring into its horizontal and vertical components. 60°Method 1: Standard PositionGiven: F = 60N, α = 60° FyFind: Fx, Fy F = 60N y Method 2: Reference AngleSolution:Fx = F cos θ x Given: F = 60N, α = 60° 60° Find: Fx, Fyθ = 360° - αθ = 360° – 60° F = 60N Solution:θ = 300° Fx = F cos α →Fx = 60 cos 300° N Fx = 60 cos 60° NFx = 30.0 N Fx = 30.0 N →Fy = F sin θ Fy = F sin α ↓Fy = 60 sin 300°N Fy = 60 sin 60° NFy = -51.96 Fy = 51.96 N ↓Fy = -52.0 N Fy = 52.0 N ↓
30. 30. Example 4: Resolve the weight of the 150 lb skater into components along the rail and normal to the rail. 1 ft 2 ft W = 150 lb
31. 31. Example 4: Solution Resolve the weight of the 150 lb skater into components along the rail and normal to the rail. 1 ft Given: W = 150 lb, rail slope= 1:2. 2 ft Find: Wx, Wy W = 150 lb Solution: x tan α = 2 ft / 1 ft y α = tan-1(2 / 1) α 1 ft α = 63.43° 2 ft Wx = W cos α  Wx α Wx = 150 cos 63.434 lb  Wy Wx = 67.1 lb  or Wx = -67.1 lb Wy = W sin α  W = 150 lb Wy = 150 sin 63.434° lb  Wy = 134.2 lb or Wy = -134.2 lb
32. 32. Exercise 3: Resolve the 75 N force on the ring into its horizontal and vertical components. 48° F = 75N
33. 33. Exercise 3: Solution Resolve the 75 N force on the ring into its horizontal and vertical components. 48° F = 75N Method 2: Reference Angle Given: F = 75 N, α = 48° y Find: Fx, Fy Solution: Fx x Fx = F cos α ← 48° Fx = 75 cos 48° N Fx = 50.2 N ← or -50.2 Fy Fy = F sin α ↓ Fy = 75 sin 48° N F = 75N Fy = 55.7 N ↓ or -55.7
34. 34. Rectangular ComponentsRectangular Components - If the magnitudes of Fx and Fy are known, the magnitude and direction of F can be calculated. y Magnitude of F: F = (Fx2 + Fy2)½ F Fy  This is the Pythagorean Theorem α θ x Fx 0 F2 = Fx2 + Fy2 Direction of F: α = tan-1Fy / Fx Given: Fx = -20 N, Fy = 5N Find: F  α is the angle of vector F with Magnitude F = (Fx2 + Fy2)½ respect to x axis (pos. or neg.). F = (400 + 25)½ = 20.6 N  To find the direction angle θ, must Direction α = tan-1Fy / Fx know which quadrant vector lies in. α = tan-15 / 20= tan-1(0.25) α = 14.04° I 0º < θ < 90º, then θ = α II 90º < θ < 180º, then θ = 180º - α 2nd quadrant since Fx is neg. III 180º < θ < 270º, then θ = 180º + α θ = 180 -14.04 = 166.0° IV 270º < θ < 360º, then θ = 360º - α θ = 166.0° or θ = - α F = 20.6 N ∠ 166.0°
35. 35. Example 5: Therectangular components of a force are given as Fx = -450 N and Fy = -300 N. Find the magnitude and direction of the force. y Fx= -450 N x 0 Fy= -300 N
36. 36. Example 5: Solution y Therectangular components of a force are given as Fx = -450 N and Fy = -300 N. Find the Fx= -450 N magnitude and direction of the force. x 0 Given: Fx = -450 N, Fy = -300 N y Find: F Fy= -300 N Magnitude – F = (Fx2 + Fy2)½ Fx= -450 N θ x F = [(-450) + (-300) ] 2 2 ½ α 0 F = 540.83 N F F = 541 N Direction- Fy= -300 N α = tan-1Fy / Fx α = tan-1300 / 450= tan-1(0.667) α = 33.7° 3rd quadrant since Fx and Fy are neg. I 0º < θ < 90º, then θ = α θ = 180 + α II 90º < θ < 180º, then θ = 180º - α θ = 180 + 33.7° III 180º < θ < 270º, then θ = 180º + α θ = 214° IV 270º < θ < 360º, then θ = 360º - α F = 541 N ∠ 214° or θ = - α
37. 37. Exercise 4: Therectangular components of a force are given as Fx = 125 N and Fy = -288 N. Find the magnitude and direction of the force. y Fx= 125 N x Fy= -288 N
38. 38. Exercise 4: Solution The y rectangular components of a force are given Fx= 125 N as Fx = 125 N and Fy = -288 N. Find the magnitude and direction of the force. x y Fy= -288 N Given: Fx = 125 N, Fy = -288 N Find: F Magnitude – F = (Fx2 + Fy2)½ Fx= 125 N θ F = [(125) + (-288) ] 2 2 ½ 0 x α F = 313.96 N F = 314 N Fy= -288 N F Direction- α = tan-1Fy / Fx α = tan-1288 / 125= tan-1(2.304) α = 66.5° 2nd quadrant since Fy is neg. I 0º < θ < 90º, then θ = α θ=-α II 90º < θ < 180º, then θ = 180º - α θ = -66.5° III 180º < θ < 270º, then θ = 180º + α F = 314 N ∠ -66.5° or 314 N ∠ 293.5° IV 270º < θ < 360º, then θ = 360º - α or θ = - α
39. 39. End Lesson 6
40. 40. Resultants by Rectangular ComponentsResultants by rectangular components: The resultant of any numberof concurrent coplanar forces can be calculated by resolving eachinto its rectangular components. y  Step 1: Resolve each force into its F1 F2 rectangular components. x Step 2: All x components are directed 0 horizontally and can be added algebraically F3 to get x component of resultant (Rx). Rx = ΣFx = (F1)x+ (F2)x + (F3)x +… Step 3: All y components are directed vertically and can be added algebraically to get y component of resultant (Ry). Ry = ΣFy = (F1)y+ (F2)y + (F3)y +… Step 4: Using x and y components of resultant calculate magnitude and direction of resultant R. R = (Rx2 + Ry2)½ α = tan-1Ry / Rx
41. 41. Example 6: Determine the resultant of the two forces using rectangular components; F1 = 3 kN ∠ 32° and F2 = 1.8 kN ∠ 105°. y F2 F1 x 0
42. 42. y Example 6: Solution F2 F1 Determine the resultant of the two forces x 0 F1 = 3 kN ∠ 32° and F2 = 1.8 kN ∠ 105°.Given: F1 = 3 kN ∠ 32° and F2 = 1.8 kN ∠ 105°.Find: Resultant R y RX Components F1 – Y Components F1 – θ2Fx1 = F1 cos θ1 Fy1 = F1 sin θ1 F2 F1Fx1 = 3000 cos 32° N Fy1 = 3000 sin 32° NFx1 = 2544.1 N Fy1 = 1589.8 N θ1 xX Components F2 – Y Components F2 – 0Fx2 = F2 cos θ2 Fy2 = F2 sin θ2Fx2 = 1800 cos 105° N Fy2 = 1800 sin 105° N Direction R –F = -465.9 N F = 1738.7 N α = tan-1Ry / RxX x2 Components R – Yy2Components R – α = tan-13328.5 / 2078.2Rx = Fx1 + Fx2 Ry = Fy1 + Fy2 α = tan-1(1.602)Rx = 2544.1 – 465.9 N Ry = 1589.8 + 1738.7 N α = 58.0°Rx = 2078.2 N Ry = 3328.5 N Rx & Ry positive so 1st quadrantMagnitude R – θ = α = 58.0°R = (Rx2 + Ry2)½ R = 3920 kN∠ 58.0°R = [(2078.2)2 + (3328.5)2]½ = 3923.9 N = 3920 N [
43. 43. Exercise 5A: Determine the resultant of the two forces using rectangular components; F1 = 150 N ∠ 28° and F2 = 100 N ∠ 99°. y F2 F1 x 0
44. 44. y Exercise 5A - Solution F2 F1 Determine the resultant of the two forces x 0 F1 = 150 N ∠ 28° and F2 = 100 N ∠ 99°.Given: F1 = 150 N ∠ 28° and F2 = 100 kN ∠ 99°.Find: Resultant R y RX Components R – θ2Rx = Fx1 + Fx2 F2 F1Rx = F1 cos θ1+ F2 cos θ2 θ1 xRx = 150 cos 28° + 100 cos 99° N 0Rx = 116.8 NY Components R –Ry = Fy1 + Fy2 Direction R –Ry = F1 sin θ1+ F2 sin θ2 α = tan-1Ry / RxRy = 150 sin 28° + 100 sin 99° N α = tan-1169.2 / 116.8Ry = 169.2 N α = 55.381° Rx & Ry positive so 1st quadrantMagnitude R –R = (Rx2 + Ry2)½ θ = α = 55.4°R = [(116.8)2 + (169.2)2]½ = 206 N [ R = 206 N∠ 55.4°
45. 45. Exercise 5B: Determine the resultant of the three forces using rectangular components; F1 = 100 N, F2 = 200 N, and F3 = 300N. y F2 40° F1 20° x 0 60° F3
46. 46. End Lesson 7
47. 47. Moment of a ForceWhat is the Moment of a Force? A force can have two effects on a rigid body;  Translation - tends to move it linearly along its line of action.  Rotation - tends to rotate it about an axis. Moment of a Force: the measure of a force’s tendency to rotate the body about an axis. The moment of a force (moment) is also called torque (Ex: torque wrench).
48. 48. Moment of a ForceWhat is the Moment of a Force? Moment of a Force in Action: Whenever there is a tendency for rotational motion to occur, a Moment is at work.What factors wouldincrease the tendencyfor rotation to occur inthe following examples?
49. 49. Moment of a Force Moment of a force : The tendency of a force to cause rotation (moment of the force) depends on two factors;  The magnitude of the force.  The perpendicular distance (d) from the center of rotation (O) to the line of action of the force. An important distinction:  Distance (d) - perpendicular distance from the center of rotation (O) to the line of action.  Distance (L) - distance from the center of rotation (O) to the point of force application. L L
50. 50. Moment of a Force Mathematical Definition of a Moment: The moment Mo of a force F about a point O is equal to the magnitude of the force multiplied by the perpendicular distance d from O to the line of action of the force. Mo = F·d Point O is referred to as the Moment Center Distance d is referred to as the Moment Arm. Units- in terms of force & distance. L  SI Units: N·m or kN·m  English Units: ft·lb or in·lb O
51. 51. Moment of a Force ALet’s test themathematicaldefinition of a momentand see its affectsusing your text book! B
52. 52. Moment of a Force Direction of a Moment: A moment is a vector just like a force is a vector. The direction of a moment vector is determined by the right hand rule. PositiveMoment - the force tends to cause a counter-clockwise rotation about the moment center . Negative Moment - the force tends to cause a clockwise rotation about the moment center . Positive Moment
53. 53. Moment of a Force Summation of Moments: In the 2D case (planar forces), moments can be added algebraically just like forces with the same line of action (Rx = F1x + F2x+ …).  Counter-clockwise  moments positive.  Clockwise  moments negative. y F1 d1 Mo = Mo1 + (-Mo2) + (-Mo3) +… d2 o xF2 d3 F3
54. 54. Moment of a Force Two ways to calculate the moment of a force about a point;Transmissibility Method: Rectangular Component Method: Extend the line of action to Find the component of the force find the moment arm length d perpendicular to rod L. (⊥ distance from center to line  Calculate the moment using; of action). Mo = Fx·d where Calculate the moment using; Fx is the normal component of F Mo = F·d where d is the moment arm = L y d is the moment arm ≠ L ϕF Fx ϕ x L L d= F d o o
55. 55. Example 7: Determine the moment of a 500 N force about point o, if θ = 30°, 60°, 90°, and 120°. θ F m m 0 20 60° o
56. 56. Example 7: Solution  Determine the moment of a  θ = 60 ° 500 N force about point o, if sin θ = d / .200 m θ = 30°, 60°, 90°, and 120°. d = .200 sin 60° mm θ F d = 0.1732 m m Mo= Fd m 0 Mo= - (500 N)(0.1732 m) 20 θ = 30 °sin θ = d / .200 m Mo= -86.6 Nm ord = .200 sin 30° m 60° Mo= 86.6 Nm d = .100 m o  θ = 90 °Mo= Fd sin θ = d / 200 mm θFMo= - (500 N)(0.1 m) d = 200 sin 90° mm m mMo= -50 Nm or d = 200 mm = 0.2 m 0 20Mo= 50 Nm  Mo= Fd d Mo= - (500 N)(0.2 m) o Mo= - 100 Nm or Mo= 100 Nm
57. 57. Example 7: Solution Determine θ F the moment of a 500 N force m m about point o, if θ = 30°, 60°, 90°, and 0 20 120°. F 60° ϕ θ o  θ = 120 °d ϕ = 180° - θ m m ϕ = 180° - 120° = 60° 0 20 sin ϕ = d / 200 mmo d = 200 sin 60° mm d = 173.2 mm = 0.1732 m Mo= Fd Mo= - (500 N)(0.1732 m)Note: Same value Mo= - 86.6 Nm oras for θ = 60º Mo= 86.6 Nm
58. 58. Exercise 6: Determine the moment of force F= 150 lb about point o given the rod length (L) is 16 inches. o L = 16 in θ=128° F= 150 lb
59. 59. Exercise 6: Solution Determine the moment of force F= 150 lb about point O given the rod length (L) is 16 inches. Given: F = 150 lb θ = 128° o L = 16 in Find: Mo L = 16 in Solution: d θ=128° ϕ = 180° - 128° = 52° ϕ d = L sin ϕ = 16 sin 52° in F= 150 lb d = 12.6082 in Mo = Fd = (150 lb)(12.61 in) Mo = 1891 in lb  Mo = (1891 in lb)(1 ft / 12 in) Mo = 157.6 ft lb 
60. 60. Moment of a Force Varignon’s Theorem: The moment of a force about any point is equal to the sum of the moments produced by the components of the force about the same point. y  Thisis the principle used in Fy Fx this method for moment calculation. In this case, the ϕ L moment due to Fy is zero. d= x F *For Moment calculations its important to remember Transmissibility Law – Force can be moved along line of action only. o Don’t move Fy tip to tail with Fx.
61. 61. Exercise 7: By Varignon’s Theorem, force F can be broken into its components acting at point A. Using this approach, find the moment about O due to the 250 N force. y A 135° x 100 mm F o B 200 mm
62. 62. Exercise 7: Solution 1 Determine the moment of the 250 N force about point O. y Given: F = 250 N; ∠ = 135° LAB= 100 mm; LBO= 200 mm 135° Find: Moment Mo about point O. A Fx x Solution: α α = 135°-90° = 45°100 mm Fy Fx = Fcos α = 250N (cos45°) F o Fx = 176.7767N B Fy = Fsin α = 250N (sin45°) 200 mm Fy = 176.7767N Mo = (0.2m)Fy – (0.1m)Fx Mo = (0.1m) 176.7767N Mo= 17.68 Nm 
63. 63. End Lesson 8
64. 64. Exercise 7: Solution 2 Determine the moment of the 250 N force about point O. Given: F = 250 N; θ = 135°; y LAB= 100 mm; LBO= 200 mm Find: Moment Mo about point O. Solution: 135° A Tan (45° +α) = LBO / LAB = 0.2 / 0.1 Fy Fx Tan (45° +α) = 2.0100 mm 45° L (45° +α) = 63.4349° α OA F o α = 18.4349° B x F = Fsin α = 250 sin 18.4349° N y 200 mm Fy = 79.0567 N Let d = L = LOA d = (0.22 + 0.12)½ = 0.22361 m Mo = Fyd = (79.0567 N)(0.2236 m) ( Mo = 17.68 Nm 
65. 65. Moment of a Force Transmissibility (review): The point of application of a force acting on a rigid body may be placed anywhere along its line of action. Example below shows how this is useful.
66. 66. Example 8: Determine the moment of the 250 N force about point O. Use the geometry method in which d (moment arm perpendicular to line of action) is determined. A 135° 100 mm F o B 200 mm
67. 67. Example 8: Solution Determine the moment of the 250 N force about Given: F = 250 N; θ = 135° point O. LAB= 100 mm; LBO= 200 mm Find: Moment Mo about point O. Solution: ϕ = 180° - 135° = 45° A 135° tan ϕ = (LBC / 100) mm LBC = (tan ϕ / 100) mm100 mm ϕ F LBC = tan 45° / 100 = 100 mm α o LCO = 200 – LBC = 100 mm B C α d α = 90° - ϕ = 45° 200 mm d = LCO sin α = 100 sin 45° mm d = 70.71 mm = 0.07071 m Mo = Fd = (250 N)(0.07071 m) Mo = 17.68 Nm 
68. 68. Exercise 8: Considering Varignon’s Theorem and Transmissibility, find the moment about O due to the 250 N force. y A 135° x 100 mm F o B 200 mm
69. 69. End Lesson 9
70. 70. Force Couples Force Couple: A couple is defined as two parallel forces with equal magnitudes but opposite sense having different lines of action separated by a distance (d).  Net force of a couple is zero, but rotates in specified direction.  Moment of a couple is not zero.  Causes rotation about an axis perpendicular to plane of its forces. A force couple is a way to produce a moment without a net force. Equivalent M = Fd
71. 71. Force Couples Moment of a Couple: The moment of a couple is given by; M = Fd The moment of a couple is independent of the choice of the moment center. MO = F (a+d) – Fa MO = Fa + Fd – Fa MO = Fd  thus MO is independent of distance a. a Themoment of a couple about any point, anywhere, is the same. OA couple may be transferred to any location in its plane and still have the same effect.
72. 72. Force Couples EquivalentCouples: Two couples acting in the same plane are equivalent if they have the same moment acting in the same direction  produce the same mechanical effect. If M = M’ then these are Equivalent Couples, and 0.4F = 0.3F’  F = (¾)F’
73. 73. Force Couples Addition of Couples: The moments about a point of two or more couples acting in a plane may be added algebraically. Using our convention ccw  = positive; cw  = negative.  Example: Find the total F2 moment about point P.P F2 d2 Mp = -(F1d1) + -(F2d2) What direction does the F1 moment act? d1 F1
74. 74. Exercise 9: Captain Bligh encounters rough seas. Determine the moment of the couple required to keep his ship on track if F= 50 lb and the wheel diameter d = 30 inches. -F F d
75. 75. Exercise 9: Solution Captain Bligh encounters rough seas. Determine the moment of the couple required to keep his ship on track if F= 50 lb and the wheel diameter d = 30 inches. Given: F = 50 lb d = 30 in Find: M Solution: M = F(d/2)  + F(d/2)  = Fd  M = (50 lb)(30 in)  M = 1500 in lb  -F F d
76. 76. Exercise 10: Determine the moment of the force couple acting on the block. Then determine the force value for an equivalent couple having opposing forces acting horizontally at C & D. Fab = 40 N C w=120 mm B h=90 mm A D Fab = 40 N
77. 77. Exercise 10: Solution  Determine the moment of Fab = 40 N the force couple acting on C the block. Then replace with w=120 mm B an equivalent couple acting horizontally at C & D. h=90 mm Given: Fab = 40 N; w = 120 mm; A h = 90 mm D Fab = 40 N Find: M and equivalent couple at C & D. Equivalent Solution: Couples M = Fabw Fcd C M = (40N)(120 mm)  B w=120 mm M = 4.80 Nm  Fcdh = M = 4.80 Nm  h=90 mm Fcd Fcd = M/h = (4.80 Nm)/(0.900 m) A D Fcd = 53.3 N (directions as shown)
78. 78. Non-Concurrent Force Systems In previous sections we;  Found the resultant of concurrent force systems.  Learned to calculate the moment of a force about a point (axis).  Learned about force couples and the moments they produce. In this section we focus on;  Non-Concurrent Force Systems .  Use what we learned about moments to find the resultant of a Non-Concurrent Force System.
79. 79. Force - Couple Systems A force system can be replaced by an equivalent system consisting of a combination of forces and couples (moments).  The net effect must be identical for each system – i.e. they must both produce the same mechanical effect.  Equivalent Force Systems – systems of forces are equivalent if; they have the same resultant force, and the same resultant moment about any selected point.  Are these Equivalent force systems? (transmissibility) Are these Equivalent force systems? (parallel force displacement)
80. 80. Force - Couple Systems  Force-Couple System – any force may be moved to another point without changing its mechanical effect, provided that an appropriate couple (moment) is added.  The value of the added couple (moment) is equivalent to the moment of the force at its original location about its new location. F F Not Equivalent d d d F F F F F F -F -F M=FdProper Method: Step 1: Step 2: Step 3:
81. 81. Example 9: Replace the 2 KN force F with a force-couple system at point B. A F = 2 KN h=390 mm B w=1.4 m
82. 82. Example 9: Solution  Replace the 2 KN force F with a force-couple system at point B. A F= 2 KN h=390 mm Given: F = 2KN h = 390 mmB w = 1.4 m w=1.4 m Find: Equivalent force-couple system A at B. Solution: F=2 KN Step 1: Step 1: Move the force to point B.B Step 2: Calculate moment about B due to F at A. A M = F(0.390 m)  M = (2000 N)(0.390 m)  F=2 KN M = 780 Nm  or -780 NmB Step 3: Step 3: Place moment at B. M=780 Nm
83. 83. Example 10: Replace the three forces shown with an equivalent force-couple system at A.F1 F2 F3
84. 84. Example 10:To find the equivalentset of forces at A. 3θ = tan   = 36.87 o −1 4Rx = ∑ Fx = 400 N cos ( 180 ) + 750 N cos ( 36.87 ) + 100 N cos ( 90 o o o ) = 200 NRy = ∑ Fy = 400 N sin ( 180o ) + 750 N sin ( 36.87 o ) + 100 N sin ( 90 o ) = 550 N
85. 85. Example 10:Find the moments about point A.Using the line of action for the force at B. The forcecan be moved along the line of action until it reachesperpendicular distance from A uuu uur r M 1 = FB d = 100 N ( 360 mm ) = 36000 N-mm
86. 86. Example 10:Find the moments about point A.The force at O can be broken up into its two componentsin the x and y direction Fx = 750 N cos ( 36.87 o ) = 600 N Fy = 750 N sin ( 36.87 o ) = 450 N Using the line of action for each component, their moment contribution can be determined.
87. 87. Example 10:Find the moments about point A.Using the line of action for Fx component d is 160 mm. uuu r uuu r M 2 = FOx d = 600 N ( 160 mm ) = 96000 N-mmFy component is 0 since inline with A.uuur uuu rMB = ∑ Mi uuu uuu uuu r r r = M1 + M 2 + M 3 r r r = 36000 N-mm k + 96000 N-mm k + 0 N-mm k r = 132000 N-mm k
88. 88. Example 10:The final result is R = 585 N at 70.0o M = 132 Nm  M = 132 Nm
89. 89. End Lesson 10
90. 90. Resultant of Nonconcurrent Coplanar Force Systems Previouslywe found the resultant (single force equivalent to the given force system) for a Concurrent, Coplanar force system – all the forces passed through a single point. Ina Nononcurrent Colanar force system;  No concurrent point exists.  Line of action for resultant is not immediately known. Q: If the forces are Nonconcurrent, how do we find the resultant? A: We need to introduce moments into our resultant formulation. Remember! Equivalent Force Systems – systems of forces are equivalent if; they have the same resultant force, and the same resultant moment about any selected point.
91. 91. Resultant of Nonconcurrent Coplanar Force SystemsSteps to find the resultant of a Nonconcurrent Coplanar force system: Step1: Find the magnitude and direction of the resultant Force. Same as before.  Choose convenient x-y coordinate system.  Resolve each force into x and y components.  Components of the reaction force are the algebraic sums of the individual force components; Rx = ΣFx Ry = ΣFy  Find magnitude of the resultant R = (Rx2+ Ry2)½  Find the direction of the resultant α = tan-1|Ry / Rx|  then find θ
92. 92. Resultant of Nonconcurrent Coplanar Force SystemsSteps to find the resultant of a Nonconcurrent Coplanar force system: Step 2: Find the location for the line of action of the resultant.  The moment of the original force system and the moment of the resultant force about an arbitrary point must be equal.  Calculate moment of the original force system about any convenient point.  Knowing this and the resultant direction & magnitude, the resultant location can be calculated.
93. 93. Example 11: Find the resultant and its location for the 2 force system shown. A F1= 9.32 KN h=300 mm B C w=1.4 m F2= 10 KN 40º
94. 94. Example 11: Solution  Find the resultant and its location for the 2 force system shown. Solution: A F= 9.32 KN Step 1: Magnitude & direction of resultant. h=300 mm Break 10KN force into components; CB Fx = 10KN cos 40º Fy = 10KN sin 40º w=1.4 m F= 10 KN Fx = 7.66 KN Fy = 6.43 KN 40º A F= 9.32 KN Find Rx = ΣFx h=300 mm Rx = (9.32 – 7.66) KN C Rx = 1.66 KN or 1.66 KN B Fx= 7.66 KN w=1.4 m Find Ry = ΣFy Fy= 6.43 KN Ry = 0 + 6.43 KN Ry = 6.43 KN or 6.43 KN
95. 95. Example 11: Solution  Find the resultant and its location for the 2 force system shown. A F= 9.32 KN h=300 mm Solution: CB Fx= 7.66 KN Find magnitude of R: w=1.4 m R = (Rx2+ Ry2)½ Fy= 6.43 KN R = [(1.662+ 6.432)]½ KN R = 6.64 KN RRy= 6.43 KN Find direction of R: α = tan-1|6.43/1.66| 75.5º α = 75.5º = θ Rx= 1.66 KN R = 6.64 KN ∠ 75.5º R= 6.64 KN 75.5º
96. 96. Example 11: Solution  Find the resultant and its location for the 2 force system shown. A h=300 mm F= 9.32 KN Solution: C Step 2: Find location of resultant.B Fx= 6.13 KN Find moment of force system about C: w=1.4 m MC = ΣMF Fy= 5.14 KN MC = (9.32KN)(300mm) + 0 + 0  y MC = 2.796 Nm  or -2.796 Nm R must produce this same moment, A so locate R accordingly. By inspection, dy R must be to the left of C to produce a C negative moment . Draw line ofB x action and solve for dx and dy which dx give MC = 2.796 Nm  or -2.796 Nm R 75.5º
97. 97. Example 11: Solution y Review data: MC = 2.796 Nm Y intercept Rx = 1.66 KN Ry = 6.43 KNX intercept A At x intercept, Rx produces no moment dy About C. C Rydx = MCB x dx dx = MC / Ry= 2.796 Nm / 6.43 KN R dx = 0.434 m 75.5º At y intercept Ry produces no moment y Rxdy = MC dy = MC / Rx= 2.796 Nm / 1.66 KN A dy = 1.684 m dy = 1.684 m θ=75.5º R θ CB x Any location along line of action is dx = 0.434 m acceptable. Choose x = -0.434 , y = 0 since located on part.
98. 98. Exercise 11: Find the resultant for the 3 force system shown and locate it with respect to point A. y F2=200 N F1=500 N 4 F3=450 N B 3 C D A x 1.5 m 1.5 m 1m
99. 99. Exercise 11: Solution  Find the resultant for the 3 force system shown and locate it with respect to point A. yRx = -300 NRy = -650 N X = 2.77 mR = 716 N 200 N 500 Nα= 65.2° 4 450 Nθ = 245° B 3 C DR = 716 N ∠ 245° A xMA= -1800 Nm 1.5 m 1.5 m 1mX = 2.77 m R = 716 N ∠ 245°
100. 100. End Lesson 11
101. 101. Resultant of Distributed Line Loads So far we’ve dealt with loads developed by individual forces concentrated at a point called Point Loads or Concentrated Loads. But what if a load is continuous along the length of a beam, say due to sacks of concrete. This continuous loading is referred to as a Distributed Load. It may be exerted along a line, over an area, or throughout an entire body (volume – remember gravity?). We will only consider the line load. Goal: determine how to represent a distributed line load as a point load resultant which we are more familiar with.
102. 102. Resultant of Distributed Line Loads LoadIntensity (w) – the magnitude of the load per unit of length over which it acts. (lb/ft) (N/m) example: 300 lb/ft - each foot of load represents 300 lb of force.Types of Distributed Loads: Uniform Load – distributed load with constant intensity (w). example: 400 N/m uniform load distributed over 2m equals 800 N total. Triangular Load – distributed load whose intensity varies linearly from 0 to some maximum value (wo). example: 300 N/m triangular load distributed over 2m equals 300 N total.
103. 103. Resultant of Distributed Line LoadsTypes of Distributed Loads: Trapezoidal Load – distributed load whose intensity varies linearly from a non-zero value to some maximum value. Can be treated as a uniform load plus a triangular load.example: a trapezoidal load with a minimum intensity of 200 N/m and a maximum intensity of 300 N/m distributed over 2m equals (200 N/m)(2 m) + (½)(100 N/m)(2m) = 500 N total.
104. 104. Resultant of Distributed Line Loads EquivalentConcentrated Force: to determine the resultant of a force system involving distributed loads, each distributed load may be replaced by its equivalent concentrated load as follows;  Magnitude – equal to area of loading diagram.  Direction – according to distributed load direction (typically vertically down).  Location – line of action passes through centroid of loading diagram.
105. 105. Resultant of Distributed Line Loads For Trapezoidal Load – break into Uniform and Triangular load first (treat them independently).
106. 106. Exercises 12: Convert each distributed load to point load(s) of appropriate magnitude and locate correctly with respect to left end of beam. 30 kN 6 kN/m B 3 kip/ft A 1 kip/ft 2m 3m 1m A B 9 ft Law 10 kips 30 kN 5 kips Rw 1 kip/ft P1 P24 kip/ft La Lb A B A B Ra A B Rb 2m 9 ft 3m 61m ft 10 ft Ra 9 ft Rb 5 ft
107. 107. Exercises 12: Solution  Convert each distributed load to a point load of appropriate magnitude and locate correctly on the beam. 30 kN 6 kN/m (6 kN/m)(5m) = 30 kN at 2.5 m A B 2m 3m 1m 3 kip/ft 1 kip/ft(1 kip/ft)(9ft) = 9 kip at 4.5kN Law 30 ft A B(½)(2 kip/ft)(9ft) = 9 kip at 6 ft Rw 9 ft A B Ra 10 kips Rb 5 kips 2m 3m 1 m kip/ft 1 P1 P2 4 kip/ft (½)(4 kip/ft)(9ft) = 18 kip at 3 ft La Lb A B (1 kip/ft)(16ft) = 16 kip at 22 ft 9 ft 6 ft 10 ft A B Ra Rb 5 ft 9 ft
108. 108. Example 12:
109. 109. Example 12: Solution
110. 110. Example 12: Solution
111. 111. Exercise 13: Determinethe resultant force of the loads acting on the beam shown, and specify its location on the beam with respect to point A.
112. 112. Exercise 13: Solution  Determine the resultant force of the loads acting on the beam shown, and specify its location on the beam with respect to point A.
113. 113. End Lesson 12
114. 114. CHAPTER TWO.. The END!
115. 115. RETIRED MATERIAL FOLLOWS:
116. 116. Exercise: Considering that the 2 force systems shown are equivalent (same mechanical effect), determine the tension in each of the equal length cables (left diagram) supporting a 1250 lb weight. I P S K Equivalent Force Systems
117. 117. Exercise: Solution Determine the tension in each of the equal length cables supporting a 1250 lb weight.For equilibrium, the resultant R from the tension (T1, T2) in the two cables must be equal and opposite force vector from the 1250 lb weight. IP 70° Solve for force triangle using law of Sines: S K 35° T2 T1 / (sin 35°) = R / (sin 110°) 110° T1 = R sin 35° / sin 110° R 35° T1 T1 = (1250)(0.5736)/(0.9397) T1 =762.98 lb 1250 lb 1250 lb T1 =763 lb T2 =763 lb
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