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Chapter 3 Rigid Bodies : Equivalent Systems of Forces Slide #: 1 Introduction In previous lessons, the bodies were assumed to be particles. Actually the bodies are a combination of a large number of particles. body particle Slide #: 2 Introduction A force may cause a body to move in a straight path that is called translation Another force may cause the body to move differently, for example, rotation The external forces may cause a motion of translation or rotation or both. The point of application of force determines the effect. Slide #: 3 Principle of Transmissibility : Equivalent Forces F F’ According to the principal of transmissibility, the effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. Two forces acting on the rigid body at two different points have the same effect on that body if they have the same magnitude, same direction, and same line of action. Two such forces are said to be equivalent. Slide #: 4 Principle of Transmissibility : Equivalent Forces • Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. Slide #: 5 Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product. • Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1.Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is V PQ sin 3. Direction of V is obtained from the right-hand rule. Slide #: 6 Vector Product of Two Vectors • Vector products: - are not commutative, Q P P Q - are distributive, P Q1 Q2 P Q1 P Q2 - are not associative, P Q S P Q S Slide #: 7 Vector Product of Unit Vectors It follows from the definition of the vector product of two vectors that the vector products of unit vectors i, j, and k are: ixi=jxj=kxk=0 i i 0 j i k k i j i j k j j 0 k j i i k j j k i k k 0 Slide #: 8 Vector Products in Terms of Rectangular Components The rectangular components of the vector product V of two vectors P and Q are determined as follows: Given P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k V=PxQ V = (Py Qz - Pz Qy ) i + (Pz Qx - Px Qz ) j + (Px Qy - Py Qx ) k V = Vx i + Vy j + Vz k Slide #: 9 Vector Products in Terms of Rectangular Components V = P x Q = Vx i + Vy j + Vz k where Vx = Py Qz - Pz Qy Vy = Pz Qx - Px Qz Vz = Px Qy - Py Qx i j k Expansion of V = P x Q = Px Py Pz Determinant Qx Qy Qz Slide #: 10 Expansion of Determinant P = Px i + Py j + Pz k y Q = Qx i + Qy j + Qz k i j k V=PxQ= Slide #: 11 Expansion of Determinant - Example P=4i+3j+2k 1 Q=6 i+1j+5k i j k V=PxQ= Slide #: 12 Calculation of Determinant : Method -1 i j k V=PxQ= 4 3 2 6 1 5 Slide #: 13 Calculation of Determinant : Method -1 4 5 6 i j k i j V=PxQ= 4 3 2 4 3 6 1 5 6 1 1 2 3 V=PxQ= 1+ 2 + 3 - 4 - 5 - 6 Slide #: 14 Calculation of Determinant : Method -1 18 k 2i 20 j i j k i j V=PxQ= 4 3 2 4 3 6 1 5 6 1 15 i 12 j 4k V = P x Q = 13 i - 8 j -14 k Slide #: 15 Calculation of Determinant : Method -2 a11 a12 a13 11 D= a21 a22 a23 21 a31 a32 a33 = D11 31 D11 : Minor of a11 in D a22 a23 D11 = = a22* a33 – a32* a23 a32 a33 Slide #: 16 Calculation of Determinant : Method -2 a11 a12 a13 11 D= a21 a22 a23 21 a31 a32 a33 = D12 31 D12 : Minor of a12 in D a21 a23 D12 = a31 a33 = a21* a33 – a31* a23 Slide #: 17 Calculation of Determinant : Method -2 a11 a12 a13 11 D= a21 a22 a23 21 a31 a32 a33 = D13 31 D13 : Minor of a13 in D a21 a22 D13 = a a = a21* a32 – a31* a22 31 32 Slide #: 18 Calculation of Determinant : Method -2 a11 a12 a13 11 D= a21 a22 a23 21 a31 a32 a33 31 D = a11 D11 - a12 D12 + a13 D13 D= a11 (a22* a33 – a32* a23) - a12 (a21* a33 – a31* a23) + a13 (a21* a32 – a31* a22) Slide #: 19 Calculation of Determinant : Method -2 i j k V=PxQ= 4 3 2 6 1 5 V=PxQ= i 3 2 - j 4 2 +k 4 3 1 5 6 5 6 1 V = P x Q = i (15-2) – j (20-12) + k (4-18) V = P x Q = 13 i – 8 j – 14 k Slide #: 20 Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. The moment of force F about point O is defined as the vector product MO = r x F where r is the position vector drawn from point O to the point of application of the force F. The angle between the lines of action of r and F is . Slide #: 21 Moment of a Force About a Point The magnitude of the moment of F about O can be expressed as MO = rF sin = Fd where d is the perpendicular distance from O to the line of action of F. • Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO. The sense of the moment may be determined by the right-hand rule. Slide #: 22 Moment of a Force About a Point • Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure. • The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane. • If the force tends to rotate the structure counter clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive. • If the force tends to rotate the structure clockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative. Slide #: 23 Varignon’s Theorem • The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O. r F1 F2 r F1 r F2 • Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F. Slide #: 24 Rectangular Components of the Moment of a Force y A (x , y, z ) Fy j yj Fx i r Fz k O xi x zk z The moment of F about O, M O r F , r xi yj zk F Fx i Fy j Fz k Slide #: 25 Rectangular Components of the Moment of a Force The rectangular components of the moment Mo of a force F are determined by expanding the determinant of r x F. i j k Mo = r x F = x y z = Mx i + My j + Mzk Fx Fy Fz where Mx = y Fz - z Fy My = zFx - x Fz Mz = x Fy - y Fx Slide #: 26 Rectangular Components of the Moment of a Force The moment of F about B, M B rA / B F rA / B rA rB x A xB i y A y B j z A z B k F Fx i Fy j Fz k i j k M B x A xB y A yB z A z B Fx Fy Fz Slide #: 27 Moment of a Force About a Point Sample Problem 3.4 SOLUTION: The moment MA of the force F exerted by the wire is obtained by evaluating the vector product, M A rC A F The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. Slide #: 28 Moment of a Force About a Point Sample Problem 3.4 SOLUTION: M A rC A F rC A AC rC rA 0.3 m i 0.08 m k F F 200 N CD CD 0.3 m i 0.24 m j 0.32 m k 200 N 0.5 m 120 N i 96 N j 128 N k i j k M A 0.3 0 0.08 120 96 128 Slide #: 29 Moment of a Force About a Point Sample Problem 3.4 SOLUTION: i j k M A 0.3 0 0.08 120 96 128 0 0.08 0.3 0.08 0.3 0 MA = i 96 - j + k -120 96 -128 -120 -128 MA = (0- (96)(0.08)) i – ((0.3)(-128) – (-120)(0.08)) j + (96)(0.3) k MA = - 7.68 i + 28.8 j + 28.8 k Slide #: 30 Moment of a Force About a Point Sample Problem 3.4 SOLUTION: M A 7.68 N m i 28.8 N m j 28.8 N mk Slide #: 31 Rectangular Components of the Moment of a Force For two-dimensional structures, M O rO A F MO F d i j k MO x y 0 Fx Fy 0 M O xFy yFx k MO MZ xFy yFx Slide #: 32 Rectangular Components of the Moment of a Force For two-dimensional structures, M B rB A F MB F d i j k M B x A xB y A yB 0 Fx Fy 0 M B x A x B Fy y A y B Fx k MB MZ x A x B Fy y A y B Fx Slide #: 33 Moment of a Force About a Point Sample Problem 3.1 A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-N vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force. Slide #: 34 Moment of a Force About a Point Sample Problem 3.1 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper. M O Fd d 24 in. cos 60 12 in. M O 100 lb 12 in. M O 1200 lb in Slide #: 35 Moment of a Force About a Point Sample Problem 3.1 b) Horizontal force at A that produces the same moment, d 24 m sin 60 20.8 m M O Fd 1200 N m F 20.8 m. 1200 N m F 20.8 m F 57.7 N Slide #: 36 Moment of a Force About a Point Sample Problem 3.1 c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA. M O Fd 1200 N m F 24 m 1200 Nm. F 24 m F 50 N Slide #: 37 Moment of a Force About a Point Sample Problem 3.1 d) To determine the point of application of vertical 240 N force to produce the same moment, M O Fd 1200 Nm 240 N d 1200 Nm. d 5m 240 N OB cos60 5 m OB 10 m Slide #: 38 Moment of a Force About a Point Sample Problem 3.1 e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force. Slide #: 39 Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is defined as P Q PQ cos scalar result • Scalar products: - are commutative, PQ Q P - are distributive, P Q1 Q2 P Q1 P Q2 - are not associative, P Q S undefined • Scalar products with Cartesian unit components, P Q Px i Py j Pz k Qx i Q y j Qz k i i 1 j j 1 k k 1 i j 0 j k 0 k i 0 P Q Px Qx Py Q y Pz Qz P P Px2 Py2 Pz2 P 2 Slide #: 40 Scalar Product of Two Vectors : Applications • Angle between two vectors: P Q PQ cos Px Qx Py Q y Pz Qz Px Qx Py Q y Pz Qz cos PQ • Projection of a vector on a given axis: POL P cos projection of P along OL P Q PQ cos PQ P cos POL Q • For an axis defined by a unit vector: POL P Px cos x Py cos y Pz cos z Slide #: 41 Mixed Triple Product of Three Vectors • Mixed triple product of three vectors, S P Q scalar result • The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign, S P Q P Q S Q S P S Q P P S Q Q P S Slide #: 42 Mixed Triple Product of Three Vectors • Evaluating the mixed triple product, S P Q S x Py Qz Pz Q y S y Pz Qx Px Qz S z Px Q y Py Qx Sx Sy Sz Px Py Pz Qx Qy Qz Slide #: 43 Moment of a Force About a Given Axis • Moment MO of a force F applied at the point A about a point O, MO r F • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis, M OL M O r F x y z M OL x y z where Fx Fy Fz Fx , Fy , Fz componentsof forceF x, y , z coordinates of point of application of F x , y , z componentsof force ( direction cosines of axis OL) Slide #: 44 Moment of a Force About a Given Axis • Moment MOL of the force F about OL measures the tendency of F to make the rigid body rotates about the fixed axis OL • Moments of F about the coordinate axes, M x yFz zFy M y zFx xFz M z xFy yFx Slide #: 45 Moment of a Force About a Given Axis If the axis does not pass through the origin • Moment of a force about an arbitrary axis, M BL M B rA B F rA B rA rB x y z M BL x A x B y A yB z A zB Fx Fy Fz Slide #: 46 Sample Problem 3.5 A cube is acted on by a force P as shown. Determine the moment of P a) about A b) about the edge AB and c) about the diagonal AG of the cube. d) Determine the perpendicular distance between AG and FC. Slide #: 47 Sample Problem 3.5 • Moment of P about A, M A rF A P rF A a i a j a i j 1 1 P P λ FC P k j- 2 2 1 1 MA a i j P j- k 2 2 MA aP i j k 2 • Moment of P about AB, M AB i MA i aP i j k 2 aP M AB 2 Slide #: 48 Sample Problem 3.5 M AG M A rA G ai aj ak rA G 1 i j k a 3 3 MA aP i j k 2 M AG 1 aP i j k i j k 3 2 aP 1 1 1 6 aP M AG 6 Slide #: 49 Sample Problem 3.5 Alternative method using determinant x y z M AG x F x A yF y A zF z A Fx Fy Fz 1/ 3 1/ 3 1/ 3 M AG a a 0 aP / 6 0 P/ 2 P/ 2 Slide #: 50 Problem 3.54 on page 105 The frame ACD is hinged at A and D. It is supported by a cable that passes through a ring at B and is attached to hooks at G and H Knowing that the tension in the cable is 1125 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable Slide #: 51 Problem 3.54 on page 105 G D A B M AD ? Slide #: 52 Problem 3.54 on page 105 G λ AD D A B rB A M AD λAD (rB A F ) AD x AD y AD z M AD x BA y BA z BA Fx Fy Fz Slide #: 53 Problem 3.54 on page 105 F F BG BG 400i 740 j 320k BG BG 400 2 740 2 320 2 BG 400i 740 j 320k BG BG 900 BG BG 0.444i 0.822 j 0.3555k BG FF BG 1125 (0.444i 0.822 j 0.3555k ) AD x AD y AD z M AD x BA y BA z BA FF BG 500 i 925 j 400 k 500 925 400 Slide #: 54 Problem 3.54 on page 105 G λ AD D A B AD 800 i 0 j 600 k AD AD 800 2 0 2 600 2 0.8 0 0.6 AD 800 i 0 j 600 k AD AD 1000 M AD x BA y BA z BA 500 925 400 AD AD 0 .8 i 0 j 0 .6 k AD Slide #: 55 Problem 3.54 on page 105 G D A B rB A x BA 400 mm 0.4m y BA 0 z BA 0 0.8 0 0.6 M AD 0.4 0 0 222 N m 500 925 400 Slide #: 56

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