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## Chapter 3 Rigid Bodies : Equivalent Systems of Forces

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**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**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.**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.**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. Principle of Transmissibility : Equivalent Forces**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: • Line of action of V is perpendicular to plane containing P and Q. • 2. Magnitude of V is • 3. Direction of V is obtained from the right-hand rule.**Vector products:**• are not commutative, • are distributive, • are not associative, Vector Product of Two Vectors**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, andk are: i xi = jxj = kxk = 0**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 = Pxi + Pyj + Pzk Q = Qxi + Qyj + Qzk V = PxQ V = (PyQz- PzQy ) i + (PzQx- PxQz ) j + (PxQy- PyQx ) k V = Vxi + Vyj + Vzk**Vector Products in Terms of Rectangular Components**V = PxQ = Vxi + Vyj + Vzk where Vx= PyQz- PzQy Vy = PzQx- PxQz Vz = PxQy- PyQx i Px Qx j Py Qy k Pz Qz Expansion of Determinant V = PxQ =**Expansion of Determinant**P = Pxi + Pyj + Pzk Q = Qxi + Qyj + Qzk Px Py Pz Qx Qy Qz i j k V = PxQ =**Expansion of Determinant - Example**P = 4 i + 3j + 2k Q = 6 i + 1j + 5 k 4 3 2 6 1 5 i j k V = PxQ =**Calculation of Determinant : Method -1**i 4 6 i 4 6 j 3 1 j 3 1 k 2 5 V = PxQ =**V = PxQ = + + - - -**1 2 3 4 5 6 Calculation of Determinant : Method -1 4 5 6 i 4 6 j 3 1 k 2 5 j 3 1 i 4 6 V = PxQ = 1 2 3**Calculation of Determinant : Method -1**18 k 2 i 20 j i 4 6 j 3 1 k 2 5 j 3 1 i 4 6 V = PxQ = 15 i 12 j 4 k V = PxQ = 13 i - 8 j -14 k**a23**a23 a22 a22 a32 a32 a33 a33 Calculation of Determinant : Method -2 a11 a11 a21 a31 a12 a12 a22 a32 a13 a13 a23 a33 D = a21 = D11 a31 D11 : Minor of a11 in D D11 = = a22* a33 – a32* a23**a23**a21 a31 a33 Calculation of Determinant : Method -2 a11 a11 a21 a31 a12 a12 a22 a32 a13 a13 a23 a33 D = a23 a21 a22 = D12 a31 a32 a33 D12 : Minor of a12 in D D12 = = a21* a33 – a31* a23**a21**a21 a22 a22 a31 a31 a32 a32 Calculation of Determinant : Method -2 a11 a11 a21 a31 a12 a12 a22 a32 a13 a13 a23 a33 D = a23 = D13 a33 D13 : Minor of a13 in D D13 = = a21* a32 – a31* a22**a21**a22 a31 a32 Calculation of Determinant : Method -2 a11 a11 a21 a31 a12 a12 a22 a32 a13 a13 a23 a33 D = a23 a33 D = a11 D11 - a12 D12+ a13 D13 D= a11 (a22* a33 – a32* a23) - a12 (a21* a33 – a31* a23) + a13 (a21* a32 – a31* a22)**3**1 4 6 2 5 3 1 2 5 4 6 i j k Calculation of Determinant : Method -2 i 4 6 j 3 1 k 2 5 V = PxQ = V = PxQ = - + V = PxQ = i (15-2) – j (20-12) + k (4-18) V = PxQ = 13 i – 8 j – 14 k**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 = rxF 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 q.**Moment of a Force About a Point**The magnitude of the moment of F about O can be expressed as MO = rF sin q = 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.**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 counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.**Varignon’s Theorem**• The moment about a given 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. • 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.**The moment of F about O,**Rectangular Components of the Moment of a Force y A (x , y, z ) Fyj yj Fxi r Fzk xi O x zk z**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. j y Fy k z Fz i x Fx Mo = rxF = = Mxi + Myj + Mzk where Mx = yFz- zFyMy = zFx- xFz Mz = xFy- yFx**Rectangular Components of the Moment of a Force**The moment of F about B,**SOLUTION:**The moment MA of the force F exerted by the wire is obtained by evaluating the vector product, Moment of a Force About a Point Sample Problem 3.4 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.**Moment of a Force About a Point Sample Problem 3.4**SOLUTION:**0**96 0.3 -120 0 96 0.08 -128 0.3 -120 0.08 -128 j k + i Moment of a Force About a Point Sample Problem 3.4 SOLUTION: - MA = 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**Moment of a Force About a Point Sample Problem 3.4**SOLUTION:**Rectangular Components of the Moment of a Force**For two-dimensional structures,**Rectangular Components of the Moment of a Force**For two-dimensional structures,**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: • moment about O, • horizontal force at A which creates the same moment, • smallest force at A which produces the same moment, • location for a 240-N vertical force to produce the same moment, • whether any of the forces from b, c, and d is equivalent to the original force.**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.**Moment of a Force About a Point Sample Problem 3.1**• Horizontal force at A that produces the same moment,**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.**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,**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.**The scalar product or dot product between two vectors P and**Q is defined as • Scalar products: • are commutative, • are distributive, • are not associative, • Scalar products with Cartesian unit components, Scalar Product of Two Vectors**Angle between two vectors:**• Projection of a vector on a given axis: • For an axis defined by a unit vector: Scalar Product of Two Vectors : Applications**Mixed triple product of three vectors,**Mixed Triple Product of Three Vectors • The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign,**Mixed Triple Product of Three Vectors**• Evaluating the mixed triple product,**Moment MO of a force F applied at the point A about a point**O, • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis, Moment of a Force About a Given Axis**Moments of F about the coordinate axes,**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**Moment of a force about an arbitrary axis,**Moment of a Force About a Given Axis If the axis does not pass through the origin**Sample Problem 3.5**A cube is acted on by a force P as shown. Determine the moment of P • about A • about the edge AB and • about the diagonal AG of the cube. • Determine the perpendicular distance between AG and FC.**Moment of P about AB,**Sample Problem 3.5 • Moment of P about A,**Sample Problem 3.5**Alternative method using determinant