Section 4.1 Moment of a Force - Scalar Formulation

Engineering Mechanics: Statics Fifteenth Edition Chapter 4 Force System Resultants Section 4.1 Moment of a Force— Scalar Formulation Section 4.2 Cross Product Section 4.3 Moment of a Force— Vector Formulation Section 4.4 Principle of Moments Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.1: Moment of a Force – Scalar Formulation (1 of 4) The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque). Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.1: Moment of a Force – Scalar Formulation (2 of 4) In a 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. In 2-D, the direction of Mo is either clockwise (CW) or counterclockwise (CCW), depending on the tendency for rotation. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.1: Moment of a Force – Scalar Formulation (3 of 4) 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. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.1: Moment of a Force – Scalar Formulation (4 of 4) Then 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. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.2: Vector Cross Product While finding the moment of a force in 2-D is straightforward when you know the perpendicular distance d, finding the perpendicular distances can be hard—especially when you are working with forces in three dimensions. So a more general approach to finding the moment of a force exists. This more general approach is usually used when dealing with three dimensional forces but can be used in the two dimensional case as well. This more general method of finding the moment of a force uses a vector operation called the cross product of two vectors. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.2: Cross Product (1 of 3) 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 As shown, uc is the unit vector perpendicular to both A and B vectors (or to the plane containing the A and B vectors). Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.2: Cross Product (2 of 3) 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 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.2: Cross Product (3 of 3) Also, the cross product can be written as a determinant. Each component can be determined using 2 × 2 determinants. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.3: Moment of a Force – Vector Formulation (1 of 2) 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. 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. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 4.3: Moment of a Force – Vector Formulation (2 of 2) So, using the cross product, a moment can be expressed as By expanding the above equation using 2  2 determinants (see Section 4.2), we get (sample units are N - m or lb - ft) MO = (ry Fz − rz Fy ) i − (rx 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. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ (1 of 2) Given: A 100 N force is applied to the frame. Find: The moment of the force at point O. Plan: 1. Resolve the 100 N force along x and y-axes. 2. Determine Mo using a scalar analysis for the two force components and then add those two moments together. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ (2 of 2) 3 +  Fy = –100   N 5 4 + → Fx = 100   N 5 Solution: Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (1 of 2) Given : F1 = { 100 i − 120 j + 75 k } lb F2 = { − 200 i + 250 j + 100 k } lb Find: Resultant moment by the forces about point O. Plan: 1. Find F = F1 + F2 and rOA. 2. Determine MO = rOA  F . Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (1 of 2) Solution: First, find the resultant force vector F F = F1 + F2 = = (100 − 200 ) i + ( −120 + −100 i + 130 j + 175 k  lb 250 ) j + (75 + 100 ) k  lb Find the position vector rOA rOA = {4i + 5 j + 3 k } ft Then find the moment by using the vector cross product. MO = i 4 −100 j 5 130 k 3 175 = [{5 (175 ) − 3 (130 )}i − {4 (175 ) − 3 ( −100 )} j + 4 (130 ) − 5 ( −100 ) k ] ft·lb = {485 i − 1000 j + 1020 k } ft·lb Copyright © 2022 Pearson Education, Inc. All Rights Reserved Concept Quiz (1 of 2) 1. If a force of magnitude F can be applied in four different 2D 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) Copyright © 2022 Pearson Education, Inc. All Rights Reserved Concept Quiz (2 of 2) 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. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving Ⅰ (1 of 2) Given: A 20 lb force is applied to the hammer. Find: The moment of the force at A. Plan: Since this is a 2-D problem: 1. Resolve the 20 lb force along the handle’s x and y axes. 2. Determine MA using a scalar analysis. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving Ⅰ (2 of 2) Solution: +  Fy = 20 sin 30° lb + → Fx = 20 cos 30° lb Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving Ⅱ (1 of 2) Given: The force and geometry shown. Find: Moment of F about point A Plan: 1. Find F and rAC . 2. Determine M A = rAC  F Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving II (2 of 2) Solution: F = { ( 80 cos 30 ) sin 40 i + ( 80 cos 30 ) cos 40 j − 80 sin30 k } N = {44.53 i + 53.07 j − 40 k } N rAC = {0.55 i + 0.4 j − 0.2 k } m Find the moment by using the cross product. i M A = 0.55 44.53 j 0.4 53.07 k − 0.2 − 40 = − 5.39 i + 13.1 j + 11.4 k } N  m Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 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 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (2 of 2) 2. If r = { 5 j } m and F = { 10 k } N, the moment r × F equals { __________ } N  m. A. 50 i B. 50 j C. –50 i D. – 50 j E. 0 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. What is the moment of the 12 N force about point A (MA )? A. 3 N  m B. 36 N  m C. 12 N  m 12 D.   N·m  3  E. 7 N·m Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 2. The moment of force F about point O is defined as M0 = ___________. A. r  F B. F  r C. r  F D. r*F Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. Particle P is in equilibrium with five (5) forces acting on it in 3-D space. How many scalar equations of equilibrium can be written for point P? A. 2 B. 3 C. 4 D. 5 E. 6 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 2. If a block is stationary, then the friction force acting on it is _______ A. ( ∑Fx ) i + ( ∑Fy ) j + ( ∑Fz ) k = 0 B.  F = 0 C.  Fx =  Fy =  Fz = 0 D. All of the above. E. None of the above. Copyright © 2022 Pearson Education, Inc. All Rights Reserved