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**Engineering Mechanics: Statics**

Chapter 4: Force System Resultants Engineering Mechanics: Statics

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Chapter Objectives To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. To provide a method for finding the moment of a force about a specified axis. To define the moment of a couple. To present methods for determining the resultants of non-concurrent force systems. To indicate how to reduce a simple distributed loading to a resultant force having a specified location.

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**Chapter Outline Moment of a Force – Scalar Formation Cross Product**

Moment of Force – Vector Formulation Principle of Moments Moment of a Force about a Specified Axis

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**Chapter Outline Moment of a Couple Equivalent System**

Resultants of a Force and Couple System Further Reduction of a Force and Couple System Reduction of a Simple Distributed Loading

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**4.1 Moment of a Force – Scalar Formation**

Moment of a force about a point or axis – a measure of the tendency of the force to cause a body to rotate about the point or axis Case 1 Consider horizontal force Fx, which acts perpendicular to the handle of the wrench and is located dy from the point O

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**4.1 Moment of a Force – Scalar Formation**

Fx tends to turn the pipe about the z axis The larger the force or the distance dy, the greater the turning effect Torque – tendency of rotation caused by Fx or simple moment (Mo) z

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**4.1 Moment of a Force – Scalar Formation**

Moment axis (z) is perpendicular to shaded plane (x-y) Fx and dy lies on the shaded plane (x-y) Moment axis (z) intersects the plane at point O

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**4.1 Moment of a Force – Scalar Formation**

Case 2 Apply force Fz to the wrench Pipe does not rotate about z axis Tendency to rotate about x axis The pipe may not actually rotate Fz creates tendency for rotation so moment (Mo) x is produced

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**4.1 Moment of a Force – Scalar Formation**

Case 2 Moment axis (x) is perpendicular to shaded plane (y-z) Fz and dy lies on the shaded plane (y-z)

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**4.1 Moment of a Force – Scalar Formation**

Case 3 Apply force Fy to the wrench No moment is produced about point O Lack of tendency to rotate as line of action passes through O

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**4.1 Moment of a Force – Scalar Formation**

In General Consider the force F and the point O which lies in the shaded plane The moment MO about point O, or about an axis passing through O and perpendicular to the plane, is a vector quantity Moment MO has its specified magnitude and direction

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**4.1 Moment of a Force – Scalar Formation**

Magnitude For magnitude of MO, MO = Fd where d = moment arm or perpendicular distance from the axis at point O to its line of action of the force Units for moment is N.m

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**4.1 Moment of a Force – Scalar Formation**

Direction Direction of MO is specified by using “right hand rule” - fingers of the right hand are curled to follow the sense of rotation when force rotates about point O

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**4.1 Moment of a Force – Scalar Formation**

Direction - Thumb points along the moment axis to give the direction and sense of the moment vector - Moment vector is upwards and perpendicular to the shaded plane

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**4.1 Moment of a Force – Scalar Formation**

Direction MO is shown by a vector arrow with a curl to distinguish it from force vector Example (Fig b) MO is represented by the counterclockwise curl, which indicates the action of F

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**4.1 Moment of a Force – Scalar Formation**

Direction Arrowhead shows the sense of rotation caused by F Using the right hand rule, the direction and sense of the moment vector points out of the page In 2D problems, moment of the force is found about a point O

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**4.1 Moment of a Force – Scalar Formation**

Direction Moment acts about an axis perpendicular to the plane containing F and d Moment axis intersects the plane at point O

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**4.1 Moment of a Force – Scalar Formation**

Resultant Moment of a System of Coplanar Forces Resultant moment, MRo = addition of the moments of all the forces algebraically since all moment forces are collinear MRo = ∑Fd taking clockwise to be positive

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**4.1 Moment of a Force – Scalar Formation**

Resultant Moment of a System of Coplanar Forces A clockwise curl is written along the equation to indicate that a positive moment if directed along the + z axis and negative along the – z axis

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**4.1 Moment of a Force – Scalar Formation**

Moment of a force does not always cause rotation Force F tends to rotate the beam clockwise about A with moment MA = FdA Force F tends to rotate the beam counterclockwise about B with moment MB = FdB Hence support at A prevents the rotation

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**4.1 Moment of a Force – Scalar Formation**

Example 4.1 For each case, determine the moment of the force about point O

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**4.1 Moment of a Force – Scalar Formation**

Solution Line of action is extended as a dashed line to establish moment arm d Tendency to rotate is indicated and the orbit is shown as a colored curl

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**4.1 Moment of a Force – Scalar Formation**

Solution

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**4.1 Moment of a Force – Scalar Formation**

Example 4.2 Determine the moments of the 800N force acting on the frame about points A, B, C and D.

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**4.1 Moment of a Force – Scalar Formation**

Solution Scalar Analysis Line of action of F passes through C

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**4.1 Moment of a Force – Scalar Formation**

Example 4.3 Determine the resultant moment of the four forces acting on the rod about point O

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**4.1 Moment of a Force – Scalar Formation**

Solution Assume positive moments acts in the +k direction, CCW View Free Body Diagram

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4.2 Cross Product Cross product of two vectors A and B yields C, which is written as C = A X B Read as “C equals A cross B”

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**4.2 Cross Product Magnitude**

Magnitude of C is defined as the product of the magnitudes of A and B and the sine of the angle θ between their tails For angle θ, 0° ≤ θ ≤ 180° Therefore, C = AB sinθ

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**4.2 Cross Product Direction**

Vector C has a direction that is perpendicular to the plane containing A and B such that C is specified by the right hand rule - Curling the fingers of the right hand form vector A (cross) to vector B - Thumb points in the direction of vector C

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4.2 Cross Product Expressing vector C when magnitude and direction are known C = A X B = (AB sinθ)uC where scalar AB sinθ defines the magnitude of vector C unit vector uC defines the direction of vector C

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**4.2 Cross Product Laws of Operations 1. Commutative law is not valid**

A X B ≠ B X A Rather, A X B = - B X A Shown by the right hand rule Cross product A X B yields a vector opposite in direction to C B X A = -C

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**4.2 Cross Product Laws of Operations 2. Multiplication by a Scalar**

a( A X B ) = (aA) X B = A X (aB) = ( A X B )a 3. Distributive Law A X ( B + D ) = ( A X B ) + ( A X D ) Proper order of the cross product must be maintained since they are not commutative

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**4.2 Cross Product Cartesian Vector Formulation**

Use C = AB sinθ on pair of Cartesian unit vectors Example For i X j, (i)(j)(sin90°) = (1)(1)(1) = 1

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**4.2 Cross Product Laws of Operations In a similar manner,**

i X j = k i X k = -j i X i = 0 j X k = i j X i = -k j X j = 0 k X i = j k X j = -i k X k = 0 Use the circle for the results. Crossing CCW yield positive and CW yields negative results

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**4.2 Cross Product Laws of Operations**

Consider cross product of vector A and B A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk) = AxBx (i X i) + AxBy (i X j) + AxBz (i X k) + AyBx (j X i) + AyBy (j X j) + AyBz (j X k) AzBx (k X i) +AzBy (k X j) +AzBz (k X k) = (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy – AyBx)k

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4.2 Cross Product Laws of Operations In determinant form,

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**4.3 Moment of Force - Vector Formulation**

Moment of force F about point O can be expressed using cross product MO = r X F where r represents position vector from O to any point lying on the line of action of F

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**4.3 Moment of Force - Vector Formulation**

Magnitude For magnitude of cross product, MO = rF sinθ where θ is the angle measured between tails of r and F Treat r as a sliding vector. Since d = r sinθ, MO = rF sinθ = F (rsinθ) = Fd

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**4.3 Moment of Force - Vector Formulation**

Direction Direction and sense of MO are determined by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F - Direction of MO is the same as the direction of the thumb

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**4.3 Moment of Force - Vector Formulation**

Direction *Note: - “curl” of the fingers indicates the sense of rotation - Maintain proper order of r and F since cross product is not commutative

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**4.3 Moment of Force - Vector Formulation**

Principle of Transmissibility For force F applied at any point A, moment created about O is MO = rA x F F has the properties of a sliding vector and therefore act at any point along its line of action and still create the same moment about O

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**4.3 Moment of Force - Vector Formulation**

Cartesian Vector Formulation For force expressed in Cartesian form, where rx, ry, rz represent the x, y, z components of the position vector and Fx, Fy, Fz represent that of the force vector

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**4.3 Moment of Force - Vector Formulation**

Cartesian Vector Formulation With the determinant expended, MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k MO is always perpendicular to the plane containing r and F Computation of moment by cross product is better than scalar for 3D problems

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**4.3 Moment of Force - Vector Formulation**

Cartesian Vector Formulation Resultant moment of forces about point O can be determined by vector addition MRo = ∑(r x F)

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**4.3 Moment of Force - Vector Formulation**

Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant Given the perpendicular distance from A to cable is rd MA = rdF In 3D problems, MA = rBC x F

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**4.3 Moment of Force - Vector Formulation**

Example 4.4 The pole is subjected to a 60N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.

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**4.3 Moment of Force - Vector Formulation**

Solution Either one of the two position vectors can be used for the solution, since MA = rB x F or MA = rC x F Position vectors are represented as rB = {1i + 3j + 2k} m and rC = {3i + 4j} m Force F has magnitude 60N and is directed from C to B

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**4.3 Moment of Force - Vector Formulation**

Solution Substitute into determinant formulation

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**4.3 Moment of Force - Vector Formulation**

Solution Or Substitute into determinant formulation For magnitude,

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**4.3 Moment of Force - Vector Formulation**

Example 4.5 Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.

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**4.3 Moment of Force - Vector Formulation**

View Free Body Diagram Solution Position vectors are directed from point O to each force rA = {5j} m and rB = {4i + 5j - 2k} m For resultant moment about O,

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**4.3 Moment of Force - Vector Formulation**

Solution For magnitude For unit vector defining the direction of moment axis,

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**4.3 Moment of Force - Vector Formulation**

Solution For the coordinate angles of the moment axis,

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