Three-Dimensional Force Systems Many problems in real-life involve 3-Dimensional Space. How will you represent each of the cable forces in Cartesian vector form? Applications Given the forces in the cables, how will you determine the resultant force acting at D, the top of the tower? 3D Rectangular Components Important: Use a righthanded coordinate system y Fy Fx z j i x k z F Fx Fy Fz F Fz y Fx i Fy j Fz k x F F F F F 2 x 2 y 2 z Direction angles & direction cosines y F b a, b and g are called the “coordinate direction angles” g a x z definition: a = angle between x-axis and F (a.k.a. = x ). b g = angle between y-axis and F (a.k.a. = y ). = angle between z-axis and F (a.k.a. = z ). more ... Determination of direction angles and direction cosines Various methods are available, including: 1) Geometry and trigonometry. y 2) If the Cartesian components of a vector are known, then: Fy F b cos a = Fx / F cos b = Fy / F cos g = Fz / F g Fz z 3) Use of vector dot product (more on this later ...) a Fx x Vector expressions using direction cosines • Direction cosines satisfy: cos2 a + cos2 bcos2 g = 1 If l = cos θx m = cos θy n = cos θz • The vector n = cos a i + cos b j + cos g k is a unit vector in the direction of F. • The vector F can be written as F = F (cos a i + cos b j + cos g k) Example 1 z A pole is subjected to a force F which has components Fx=1.5 kN and Fz=1.25 kN. If b = 75°, determine the magnitudes of F and Fy. Answer: F = 2.02 kN Fy = 0.523 kN Fz a Fx x g F b Fy y Example 2 Determine the magnitude and direction angles for the resultant force. A: Rx = 348 N Ry = 75 N Rz = -97 N a = 19.4° b = 78.3° g = 105° z F2 = 250 N 5 3 4 y 30 60 60 x 45 F1 = 350 N z F2 = 250 N 5 F2z 3 F2a F2y a x 4 30 y F2x F1 = 350 N Relative Position Vectors: The position of a point B relative to another point A: B(x , y , z ) head B B B A(xA, yA, zA) tail r = (xB-xA) i +(yB-yA) j + (zB-zA) k z B (xB -xA )i A x r (zB -zA )k y (yB -yA )j Position vector components = coordinates of head - coordinates of tail. If the coordinates of points A and B are known, the force F can be written as Example 3 Express the position vector r in Cartesian vector form. Then determine the magnitude and coordinate direction angles. z 1m y 1m 2m x 4m r 4m 1m A: r = (-3i - 6j + 2k) m, a = 115.4°, b = 149°, g = 73.4° Force vectors directed along a line Many of the structural members we will deal with will support forces that are directed along the axis of the member (e.g., cables, ropes, bars, often times columns, ...). Thus, by knowing the orientation of the member (position vector), we can conveniently express its force in vector form. Procedure: 1) Determine the position vector r for the member. 2) Compute a unit vector along the axis of the member 3) The member’s force vector is then F= Fn. n = r/r. Example 4 Determine the x,y and z components of the force vector shown. Also, determine the direction angles. y 25 600 N x 30 z A: F = (220i + 544j + 127k) N, a = 68.5°, b = 25.0°, g = 77.8° Example 5: A section of pre-cast concrete wall is temporarily held by the cables shown. Knowing that the tensile forces in cables AB and AC are 840 N and 1200 N, respectively, determine the magnitude and direction of the resultant force applied to the stake at A. C D B 8m A 11 m 16 m A: R = (-1440i + 720j - 360k) N Example 6: Determine the magnitude of the component of the force F along the Aa axis. Answer: FAa = 36.0 N Dot Product - Applications For this geometry, can you determine angles between the pole and the cables? For force F at Point A, what component of it (F1) acts along the pipe OA? What component (F2) acts perpendicular to the pipe? Dot Product - Definition The dot product of vectors A and B is defined as A•B = A B cos θ. Angle θ is the smallest angle between the two vectors and is always in a range of 0º to 180º. Dot Product Characteristics: 1. The result of the dot product is a scalar (a positive or negative number). 2. The units of the dot product will be the product of the units of the A and B vectors. Examples: i•j=0 i•i=1 A • B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k) = Ax Bx + AyBy + AzBz Example Given: The force acting on the pole Find: The magnitude of the projection of the force along the pole OA. A rOA = {2 i + 2 j – 1 k} m rOA = (22 + 22 + 12)1/2 = 3 m nOA = 2/3 i + 2/3 j – 1/3 k F = {2 i + 4 j + 10 k}kN FOA = F • nOA = (2)(2/3) + (4)(2/3) + (10)(-1/3) = -2/3 kN Moment in 3-D Vector Formulation Using the vector cross product, MO = r × F . Here r is the position vector from point O to any point on the line of action of F. Principle of Moments Consider F = F1 + F2 Moment of F about point O MO = = r ×F = r ×(F1 + F2 ) = r ×F1 + r ×F2 The moment of a force about a point is equal to the sum of the moment of the force’s components about the point. Cross Product nC In general, the cross product of two vectors A and B results in another vector C , i.e., C = A ×B. The magnitude and direction of the resulting vector can be written as C = A × B = A B sin θ nC Laws of operation special commutative law AxB=-BxA scalar multiplication a (A x B ) = (a A) x B = A x (a B) distributive law Ax(B+C)=AxB + AxC Cross product between Cartesian vectors Let A = Ax i + Ay j + Az k and B = Bx i + By j +Bz k Then AxB = (Ay Bz - Az By ) i (Ax Bz - Az Bx ) j + (Ax By - Ay Bx ) k more ... ... or ... computation of cross product by expansion of determinant i A B det Ax B x A B Ay Bzi Az Bx j Ax By k - (Ay Bx k Az Byi Ax Bz j) j Ay By k i Az Ax Bz Bx j Ay k i Az Ax j Ay By Bz Bx By ... or ... expand determinat using co-factors i A B det Ax Bx . . = i . Ay . By j Ay By . . Az - j Ax Bz Bx k Az Bz . . . . Az k Ax . Bz Bx . . Ay . By . i(A B B A ) j( A B B A ) k( A B B A ) y z y z x z x z x y x y Cross Product The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product. For example: i × j = k Note that a vector crossed into itself is zero, e.g., i × i = 0 z demonstration of orthogonality properties: k x Let A = i and B = j. Then j y i i j i j det 1 0 0 1 k 0 k ! 0 Similarly: ixi=0 j x i = -k kxi=j ixj=k jxj=0 k x j = -i i x k = -j jxk=i kxk=0 more ... Demonstration of i x j = k using C = A x B = (A B sin ) uC : Magnitude of the cross product is | i x j | = (1)(1) sin 90° = 1. The direction of the cross product result C is: z C x j y i Thus, we observe that C = k. Example Determine the magnitude of the moment of the 600 N force about A. Solution rAB = (-0.6i – 0.5j + 0.4k) m MA = rAB x F = (-0.6i – 0.5j + 0.4k) x (600i) MA = 240 j + 300 k N.m MA = 384 N.m Moment of a force about an axis Our goal is to find the moment of F (the tendency to rotate the body) about the axis a’-a. First compute the moment of F about any arbitrary point O that lies on the a’- a axis using the cross product. MO = r × F Moment of a force about an axis Now, find the component of Ma along the axis a’-a using the dot product. Ma = na • MO Ma can also be obtained as na x M a na r F rx Fx na y ry Fy na z rz Fz The above equation is also called the triple scalar product. Moment of a force about an axis Ma = na • ( r ×F ) na represents the unit vector along the axis a’-a axis, r is the position vector from any point on the a’-a axis to any point A on the line of action of the force, and F is the force vector. Example Given: A force is applied to the tool to open a gas valve. Find: The magnitude of the moment of this force about the z axis of the valve. Solution n=1k rAB = {0.25 sin 30° i + 0.25 cos30° j} m = {0.125 i + 0.2165 j} m F = {-60 i + 20 j + 15 k} N Mz = n • ( rAB × F ) 0 0 M z 0.125 0.2165 - 60 20 1 0 = 1{0.125(20) – 0.2165(-60)} N·m = 15.5 N·m 15 Example The magnitude of the force F is 0.2 N and its direction cosines are cos θx = 0.727, cos θy = -0.364, and cos θz = 0.582. Determine the magnitude of the moment of F about the axis AB of the spool. Solution rAB = (0.3i – 0.1j – 0.4k) m rAB 0.32 0.12 0.42 0.26 m nAB = (0.3i – 0.1j – 0.4k) / (0.26)1/2 m F = 0.2 (0.72i – 0.364j – 0.582k) m rAP = (0.26i – 0.025j – 0.11k) m i M A 0.26 j k - 0.025 - 0.11 0.727 - 0.364 0.582 Solution MAB = MA . nAB Example What is the moment of the force F in figure about the bar BC? Solution Determine a Vector r: We need a vector from any point on the line BC to any point on the line of action of the force. We can let r be the vector from B to the point of application of F: rBA = (4 - 0)i + (2 - 0)j + (2 - 3)k = 4i + 2j - k (m) nBC Solution Determine a Vector n: To obtain a unit vector along the bar BC, we determine the vector from B to C: (0 - 0)i + (4 - 0)j + (0 - 3)k = 4j - 3k (m) Divide it by its magnitude: n BC 4 j - 3k (m) 4 m - 3 m 2 2 0.8 j - 0.6k Solution Evaluate MBC (in scalar form): 0 n BC r F 4 -2 0.8 - 0.6 2 - 1 -24.8 kN - m 6 3 the moment of F about the bar BC in vectorial form: MBC = [nBC (r F)]nBC = -24.8 nBC Couples • It is possible to exert a moment on an object without subjecting it to a net force: – E.g. when a compact disk begins rotating or a screw is turned by a screwdriver – Forces are exerted on these objects in such a way that the net force is zero while the net moment is not zero Couples • Couple: 2 forces that have equal magnitudes, opposite directions & different lines of action – Tends to cause rotation of an object even though the vector sum of the forces is zero & has the same remarkable property that the moment it exerts is the same about any point Couples • The moment of a couple is simply the sum of the moments of the forces about a point P: M = [r1× F] + [r2 × (-F)] = (r1 - r2) × F – The vector r1 - r2 is equal to the vector r: M=r×F – Since r doesn’t depend on the position of P, the moment M is the same for any point P • A couple is often represented in diagrams by showing the moment: Couples • Notice that M = r × F is the moment of F about a point on the line of action of -F: – The magnitude of the moment of a force about a point equals the product of the magnitude of the force & the perpendicular distance from the point to the line of action of the force: |M| = D|F| where D is the perpendicular distance between the lines of action of the 2 forces Couples – The cross product r × F is perpendicular to r & F, which means that M is perpendicular to the plane containing F & -F – Pointing the thumb of the right hand in the direction of M, the arc of the fingers indicates the direction of the moment Couples • A plane containing 2 forces is perpendicular to our view: – The distance between the lines of action of the forces is 4 m, so the magnitude of the moment of the couple is: |M| = (4 m)(2 kN) = 8 kN-m Couples – The moment M is perpendicular to the plane containing the 2 forces: • Pointing the arc of the fingers of the right hand counterclockwise, the right-hand rule indicates that M points out of the page • Therefore, the moment of the couple is: M = 8k (kN-m) – We can also determine the moment of the couple by calculating the sum of the moments of the 2 forces about any point Couples • The sum of the moments of the forces about the origin O is: M = [r1× (2j)] + [r2 × (-2j)] = [(7i + 2j) × (2j)] + [(3i + 7j) × (-2j)] = 8k (kN-m) Couples – In a 2-D situation like this example, we represent the couple by showing its magnitude & a circular arrow that indicates its direction: • It is not convenient to represent the couple by showing the moment vector because the vector is perpendicular to the page Couples • By grasping a bar & twisting it, a moment can be exerted about its axis: – Although the system of forces exerted is distributed over the surface of the bar in a complicated way, the effect is the same as if 2 equal & opposite forces are exerted Couples – When we represent a couple as in the figure or by showing the moment vector M, we imply that some system of forces exerts that moment: – The system of forces is nearly always more complicated than 2 equal & opposite forces but the effect is the same: • Model the actual system as a simple system of 2 forces Example The force F in Figure is 10i - 4j (N). Determine the moment of the couple & represent it as shown in next figure Example Strategy We can determine the moment in 2 ways: 1st Method: Calculate the sum of the moments of the forces about a point 2nd Method: Sum the moments of the 2 couples formed by the x & y components of the forces Example Solution 1st Method: If we calculate the sum of the moments about a point on the line of action of 1 of the forces, the moment of that force is zero & we only need to calculate the moment of the other force. Choosing the point of application of F, the moment is: M = r × (-F) = (-2i + 3j) × (-10i + 4j) = 22k (N-m) Example Solution 2nd Method: The x & y components of the forces from 2 couples. Determine the moment of the original couple by summing the moments of the couples formed by the components: Example Solution Consider the 10-N couple. The magnitude of its moment is (3 m)(10 N) = 30 N-m & its direction is counterclockwise, indicating that the moment vector points out of the page. Therefore, the moment is 30k Nm. The 4-N couple causes a moment of magnitude (2 m)(4 N) = 8 N-m & its direction is clockwise, so the moment is -8k N-m. Example Solution The moment of the original couple is: M = 30k - 8k = 22k (N-m) Its magnitude is 22 N-m & its direction is counterclockwise: Example - Sum of the Moments Due to 2 Couples Determine the sum of the moments exerted on the pipe in figure by the 2 couples. Example Strategy Express the moment exerted by each couple as a vector: To express the 30-N couple in terms of a vector, express the forces in terms of their components. Then sum the moment vectors to determine the sum of the moments exerted by the couples. Example Solution Consider the 20-N couple. The magnitude of the moment of the couple is: (2 m)(20 N) = 40 N-m The direction of the moment vector is perpendicular to the y-z plane & the right-hand rule indicates that it points in the positive x axis direction. The moment of the 20-N couple is 40i (N-m). Example Solution By resolving the 30-N forces into y & z components, we obtain the 2 couples: Example Solution The moment of the couple formed by the y components is -(30 sin 60°)(4)k (N-m) & the moment of the couple formed by the z components is (30 cos 60°)(4)j (N-m). The sum of the moments is therefore: Σ M = 40i + (30 cos 60°)(4)j - (30 sin 60°)(4)k (N-m) = 40i + 60j - 104k (N-m) Example The two forces acting on the handles of the pipe wrenches constitute a coupe M. Express the couple vector. Answer: M = -75 i + 22.5 j N.m Extra Question 1 The tension in cable AB is 2 kN. What is the magnitude of the moment about shaft CD due to the force exerted by the cable at A? Answer: MCD = 1.633 kN.m Extra Question 2 Determine the distance d between A and B so that the resultant couple moment has a magnitude of MR = 20 N.m Extra Question 3 Determine the total moment on point O. Resultant Example: Representing a Force by a Force and Couple System 1 in Figure consists of a force FA = 10i + 4j - 3k (N) acting at A. Represent it by a force acting at B and a couple. Example Strategy Represent the force FA by a force F acting at B & a couple M (system 2). Determine F & M by using the 2 conditions for equivalence. Example Solution The sums of the forces must be equal: (Σ F)2 = (Σ F)1: F = FA = 10i + 4j - 3k (N) The sums of the moments about an arbitrary point must be equal: The vector from B to A is: rBA = (4 - 8)i + (4 - 0)j + (2 - 6)k = -4i + 4j - 4k (m) Example Solution So the moment about B in system 1 is: i j k rBA FA - 4 4 - 4 4i - 52 j - 56k (N - m) 10 4 - 3 The sums of the moments about B must be equal: (MB)2 = (MB)1: M = 4i - 52j - 56k (N-m) Example The building slab has four columns. F1 and F2 = 0. Find the equivalent resultant force and couple moment at the origin O. Also find the location (x,y) of the single equivalent resultant force. Solution FRO = {-50 k – 20 k} = {-70 k} kN MRO = (10 i) × (-20 k) + (4 i + 3 j)x(-50 k) = {200 j + 200 j – 150 i} kN·m = {-150 i + 400 j } kN·m The location of the single equivalent resultant force is given as, x = -MRyo/FRzo = -400/(-70) = 5.71 m y = MRxo/FRzo = (-150)/(-70) = 2.14 m

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