Engineering Mechanics : STATICS BDA10203 Lecture #09 Group of Lecturers Faculty of Mechanical and Manufacturing Engineering Universiti Tun Hussein Onn Malaysia

MOMENT OF A FORCE (Section 4.1) Today’s Objectives : Students will be able to: a) understand and define moment, and, b) determine moments of a force in 2-D and 3-D cases. Learning Topics : Applications Moment in 2-D Moment in 3-D

APPLICATIONS (continued) What is the effect of the 30 N force on the lug nut?

MOMENT IN 2-D The 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 Sometimes called a torque

When apply force Fy to the wrench No moment is produced about point O MOMENT IN 2-D Note that When 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

MOMENT IN 2-D (continued) Moment MO is vector quantity and has its specified magnitude and direction In the 2-D case, the magnitude of the moment is Mo = F d As shown, d is the perpendicular distance from point O to the line of action of the force, F. Units for moment is N.m

MOMENT IN 2-D (continued) In 2-D, the direction of MO is either clockwise or counter-clockwise depending on the tendency for rotation. 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 -Using the right hand rule, the direction and sense of the moment vector points out of the page The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive

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

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 The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. Mo +

Solution

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

Solution Line of action of F passes through C The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. Mx +

RESULTANT MOMENT IN 2-D Resultant moment of a system of coplanar forces can be determined by simply adding the moments of all forces algebraically since all the moment vectors are collinear. + MRo = ∑Fd

Determine the resultant moment of the four Example 4.3 Determine the resultant moment of the four forces acting on the rod about point O Solution Assume positive moments acts in the +k direction, CCW

MOMENT IN 3-D Moments in 3-D can be calculated using scalar (2-D) approach but it can be difficult and time consuming. Thus, it is often easier to use a mathematical approach called the vector cross product. Read as “Momen equals vector r cross vector F”

CROSS PRODUCT 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  UC Here UC is the unit vector perpendicular to both A and B vectors as shown (or to the plane containing the A and B vectors).

Cartesian Vector Formulation 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 For i X i, (i)(i)(sin0°) = (1)(1)(0) = 0

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 , j x i = -k Note that a vector crossed into itself is zero, e.g., i  i = 0

Consider cross product of vector A and B 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) A X B = (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy – AyBx)k

CROSS PRODUCT (continued) Of even more utility, the cross product can be written as Each component can be determined using 2  2 determinants.

MOMENT IN 3-D 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

Magnitude Direction MOMENT IN 3-D For magnitude of cross product, MO = r F 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 Direction - “curl” of the fingers indicates the sense of rotation

MOMENT IN 3-D (continued) So, using the cross product, a moment can be expressed as By expanding the above equation using 2  2 determinants we get (sample units are N - m) MO = (r y FZ - rZ Fy) i - (r x Fz - rz Fx ) j + (rx Fy - ry Fx ) k The physical meaning of the above equation becomes evident by considering the force components separately and using a 2-D 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 MO = rA x F = rB x F = rC x F

PRINCIPLE OF TRANSMISSIBILITY 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, using cross product, MA = rAB x F = rAC x F

RESULTANT MOMENT IN 3-D MRo = ∑(r x F) Resultant moment of a System of Forces : If a body is acted upon by a system of forces as shown in the figure, the resultant moment of the forces about point O can be determined by vector addition as MRo = ∑(r x F)

EXAMPLE 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. Solution Either one of the two position vectors can be used for the solution, MA = rB x F or MA = rC x F

Position vectors are represented as rB = {1i + 3j + 2k} m and Solution 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 Substitute into determinant formulation

Solution Or For magnitude,

EXAMPLE Given: The pole support a 100N (≈ 10kg) traffic light. Find: Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A. Plan: 1) Find a position vector of traffic light. 2) Using Cartesian vector notation, determine the force that the traffic light acted on the pole. 3) MA = r X F 5.4m=

EXAMPLE (continued) F = F (UCA) Vector = Magnitude x direction F = F (UCA) UCA = rCA / | rCA | = -5.4 k / 5.4 = -1 k F = 100{-1k} = -100k B 5.4m= A(0,0,0) F rAD = {( XD – XA)i +(YD – YA)j +(ZD – ZA) k}m = {(3.6sin30 – 0)i + (3.6cos30 – 0)j + (5.4 – 0)k}m = {1.8i + 3.12j + 5.4k}m D (3.6 sin 30, 3.6 kos 30, 0) MA = rAD X F = {(1.8i + 3.12j + 5.4k) x (-100k)} Nm = (3.12 x -100)i - (1.8 x -100)j + (0) k = -312 i + 180 j

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: The man pulls on the rope with a force of F = 20N. Find: Determine the moment that this force exerts about the base of the pole at O. Plan: 1) Find a position vector of rOA and rAB. 2) Using Cartesian vector notation, determine the force that acted on the pole. 3) MA = r X F

GROUP PROBLEM SOLVING (Continue) rAB = {( XB – XA)i +(YB – YA)j +(ZB – ZA) k}m = {(4 – 0)i + (-3 – 0)j + (1.5 – 10.5)k}m = {4i - 3j - 9k}m rOA = {( XA – XO)i +(YA – YO)j +(ZA – ZO) k}m = {(0 – 0)i + (0 – 0)j + (10.5 – 0)k}m = {0i + 0j + 10.5k}m F = FuAB = F(rAB /rAB ) = 20{0.4i - 0.3j - 0.9k} ={8i – 6j - 18k} N MO = rOA X F = {63i + 84j} Nm

HOMEWORK TUTORIAL Q1 (4-39): The curved rod lies in the x-y plane and has a radius r. If a force F acts at its end as shown, determine the moment of this force about point B.

HOMEWORK TUTORIAL (continued) Q2 (4-40): The force F acts at the end of the beam. Determine the moment of the force about point A.

Resultant Moment of Forces 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.

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,

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

Solution For the coordinate angles of the moment axis,

PRINCIPLE OF MOMENTS F a b d O For example, MO = F d and the direction is counter-clockwise. Often it is easier to determine MO by using the components of F as shown. a b O F F x F y Using this approach, MO = (FY a) – (FX b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive.

PRINCIPLE OF MOMENTS Principle of moments is also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point” For F = F1 + F2, MO = r X F1 + r X F2 = r X (F1 + F2) = r X F

Principles of Moments Example 1.0 A 200N force acts on the bracket. Determine the moment of the force about point A.

Principles of Moments Solution Method 1: From trigonometry using triangle BCD, CB = d = 100cos45° = 70.71mm = 0.07071m Thus, MA = Fd = 200N(0.07071m) = 14.1N.m (CCW) As a Cartesian vector, MA = {14.1k}N.m

Principles of Moments Solution Method 2: Resolve 200N force into x and y components Principle of Moments MA = ∑Fd MA = (200sin45°N)(0.20m) – (200cos45°)(0.10m) = 14.1 N.m (CCW) Thus, MA = {14.1k}N.m

Principles of Moments Example 2.0 The force F acts at the end of the angle bracket. Determine the moment of the force about point O.

Principles of Moments Solution Method 1 MO = 400sin30°N(0.2m)-400cos30°N(0.4m) = -98.6N.m = 98.6N.m (CCW) As a Cartesian vector, MO = {-98.6k}N.m

Principles of Moments Solution Method 2: Express as Cartesian vector r = {0.4i – 0.2j}N F = {400sin30°i – 400cos30°j}N = {200.0i – 346.4j}N For moment,

EXAMPLE 3.0 Given: A 400 N force is applied to the frame and  = 20°. Find: The moment of the force at A. Plan: 1) Resolve the force along x and y axes. 2) Determine MA using scalar analysis.

EXAMPLE (continued) Solution +  Fy = 400 sin 20° N +  Fx = 400 cos 20° N + MA = {(400 sin 20°)(3) + (400 cos 20°)(2)} N·m = 1162 N·m

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: A 40 N force is applied to the wrench. Find: The moment of the force at O. Plan: 1) Resolve the force along x and y axes. 2) Determine MO using scalar analysis. Solution: Fy = 40 cos 20° N Fx = 40 sin 20° N + MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = -7107 N·mm = - 7.11 N·m

ATTENTION QUIZ 10 N 5 N 3 m P 2 m 1. Using the CCW direction as positive, the net moment of the two forces about point P is A) 10 N ·m B) 20 N ·m C) - 20 N ·m D) 40 N ·m E) - 40 N ·m 2. If r = { 5 j } m and F = { 10 k } N, the moment r x F equals { _______ } N·m. A) 50 i B) 50 j C) –50 i D) – 50 j E) 0 Answers: 1. B 2. A

CONCEPT QUIZ 1. If a force of magnitude F can be applied in four different 2-D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum torque values on the nut. (Max, Min). A) (Q, P) B) (R, S) C) (P, R) D) (Q, S) Answers: 1. D 2. A 2. If M = r  F, then what will be the value of M • r ? A) 0 B) 1 C) r 2 F D) None of the above.

HOMEWORK TUTORIAL Q1 (4-4): Determine the magnitude and directional sense of the resultant moment of the forces A and B about point O. =45° =3m =5m =30° =40kN 3m= =13m =6m =60kN (Ans: 858 kN.m CW)

HOMEWORK TUTORIAL (continued) Q2 (4-5): Determine the magnitude and directional sense of the resultant moment of the forces A and B about point P. =45° =3m =5m =30° =40kN 3m= =13m =6m =60kN (Ans: 1165 kN.m CW)

HOMEWORK TUTORIAL (continued) Q3 (4-31): The worker is using the bar to pull two pipes together in order to complete the connection. If he applies a horizontal force F to the handle of the lever, determine the moment of this force about the end A. What would be the tension T in the cable needed to cause the opposite moment about point A. Ans: MA=460N.m CCW T = 3.26kN