FORCE VECTORS, VECTOR OPERATIONS & ADDITION COPLANAR FORCES Objective: a) Resolve a 2-D vector into components. b) Add 2-D vectors using Cartesian vector notations. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. An example of the process: Break the three vectors into components, then add them. FR = F1 + F2 + F3 = F1x i + F1y j − F2x i + F2y j + F3x i − F3y j = (F1x − F2x + F3x) i + (F1y + F2y − F3y) j = (FRx) i + (FRy) j Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. You can also represent a 2-D vector with a magnitude and angle. FR = FRx2 + FRy2  = tan Statics, Fourteenth Edition R.C. Hibbeler −1 FRy FRx Copyright ©2016 by Pearson Education, Inc. All rights reserved. EXAMPLE I Given: Three concurrent forces acting on a tent post. Find: The magnitude and angle of the resultant force. Plan: a) Resolve the forces into their x-y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. EXAMPLE I (continued) F1 = {0 i + 300 j } N F2 = {– 450 cos (45°) i + 450 sin (45°) j } N = {– 318.2 i + 318.2 j } N F3 = { (3/5) 600 i + (4/5) 600 j } N = { 360 i + 480 j } N Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. EXAMPLE I (continued) Summing up all the i and j components respectively, we get, FR = { (0 – 318.2 + 360) i + (300 + 318.2 + 480) j } N = { 41.80 i + 1098 j } N y Using magnitude and direction: FR FR = ((41.80)2 + (1098)2)1/2 = 1099 N  = tan-1(1098/41.80) = 87.8°  x Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. GROUP PROBLEM SOLVING Given: Three concurrent forces acting on a bracket. Find: The magnitude and angle of the resultant force. Show the resultant in a sketch. Plan: a) Resolve the forces into their x and y-components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. GROUP PROBLEM SOLVING (continued) F1 = {850 (4/5) i − 850 (3/5) j } N = { 680 i − 510 j } N F2 = {- 625 sin (30°) i − 625 cos (30°) j } N = {- 312.5 i − 541.3 j } N F3 = {-750 sin (45°) i + 750 cos (45°) j } N {- 530.3 i + 530.3 j } N Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. GROUP PROBLEM SOLVING (continued) Summing all the i and j components, respectively, we get, FR = { (680 − 312.5 − 530.3) i + (−510 − 541.3 + 530.3) j }N = {− 162.8 i − 520.9 j } N Now find the magnitude and angle,  FR = ((− 162.8)2 + (− 520.9)2) ½ = 546 N  = tan–1( 520.9 / 162.8 ) = 72.6° From the positive x-axis,  = 253° Statics, Fourteenth Edition R.C. Hibbeler -162.8 y x  FR Copyright ©2016 by Pearson Education, Inc. All rights reserved. -520.9 Cartesian Unit Vectors For a vector A, with a magnitude of A, an unit vector is defined as uA = A . A Characteristics of a unit vector : a. Its magnitude is 1. b. It is dimensionless (has no units). c. It points in the same direction as the original vector (A). The unit vectors in the Cartesian axis system are i, j, and k. They are unit vectors along the positive x, y, and z axes respectively. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Direction of a Cartesian Vector (1 of 2) The direction or orientation of vector A is defined by the angles ,, and . These angles are measured between the vector and the positive X, Y and Z axes, respectively. Their range of values are from 0° to 180° Using trigonometry, “direction cosines” are found using Ay Ax A cos = cos = cos = z A A A These angles are not independent. They must satisfy the following equation. cos² + cos² + cos²  = 1 Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Direction of a Cartesian Vector (2 of 2) This result can be derived from the definition of a coordinate direction angles and the unit vector. Recall, the formula for finding the unit vector of any position vector: uA = or written another way, Statics, Fourteenth Edition R.C. Hibbeler A A A Ax = i+ y j+ z k A A A A u A = cos  i + cos  j + cos  k . Copyright ©2016 by Pearson Education, Inc. All rights reserved. Position Vectors & Force Vectors Today’s Objectives: a. Represent a position vector in Cartesian coordinate form, from given geometry. b. Represent a force vector directed along a line. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Position Vector (1 of 2) A position vector is defined as a fixed vector that locates a point in space relative to another point. Consider two points, A and B, in 3-D space. Let their coordinates be (X A, Y A, Z A) and (X B, Y B, Z B), respectively. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (1 of 3) Given: The 420 N force along the cable AC. Find: The force F A C in the Cartesian vector form. Plan: 1. Find the position vector rAC and its unit vector uAC 2. Obtain the force vector as FA C = 420 N uAC. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (2 of 3) As per the figure, when relating A to C, we will have to go 2 m in the x-direction, 3 m in the y-direction, and −6 m in the z-direction. Hence, rAC = {2i + 3 j − 6k } m. (We can also find rAC by subtracting the coordinates of A from the coordinates of C.) Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (3 of 3)  rAC = 22 + 32 + ( −6 ) 2  =7m 1 2 r r  Now uAC = rAC and FAC = 420 uAC = 420  AC   rAC  AC So FAC = 420  (2i + 3 j − 6k ) N  7  = 120i +180 j −360k N Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (1 of 3) Given: Two forces are acting on a flag pole as shown in the figure. FB = 560 N and FC = 700 N Find: The magnitude and the coordinate direction angles of the resultant force. Plan: 1. Find the forces along A B and A C in the Cartesian vector form. 2. Add the two forces to get the resultant force, FR . 3. Determine the magnitude and the coordinate angles of Statics, Fourteenth Edition R.C. Hibbeler FR . Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (2 of 3) rAB = 2i − 3 j − 6k  m rAC = 3i + 2 j − 6k  m  = 3 rAB = 22 + ( −3 ) + ( −6 ) rAC 2 2 + 22 + ( −6 ) 2  1 2  1 2 2 =7m =7m   FAB = 560  rrAB  N FAC =700 rAC  N FAB = 560( 2i – 3 j – 6k ) FAC = 700(3i + 2 j – 6k ) FAB = (160i –240 j – 480k ) N FAC =300i + 200 j –600k  N  AB  7N Statics, Fourteenth Edition R.C. Hibbeler  rAC  7N Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (3 of 3) FR = FAB + FAC = 460i –40 j – 1080k  N  FR = 4602 + ( −40 ) + ( −1080 ) 2 2  1 2 = 1174.6 N FR = 1175 Ν  = cos−1 460  o = 66.9   1175  −40  o = 92.0   1175   = cos−1 −1080  o = 157   1175   = cos−1 Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Equilibrium of A Particle, The Free-body Diagram & Coplanar Force Systems Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. The What, Why and How of A Free Body Diagram (FB D) (1 Of 2) Free-body diagrams are one of the most important things for you to know how to draw and use for statics and other subjects! What? - It is a drawing that shows all external forces acting on the particle. Why? - It is key to being able to write the equations of equilibrium— which are used to solve for the unknowns (usually forces or angles). Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. The What, Why and How of A Free Body Diagram (FBD) (2 of 2) How? 1. Imagine the particle to be isolated or cut free from its surroundings. 2. Show all the forces that act on the particle. Active forces: They want to move the particle. Reactive forces: They tend to resist the motion. 3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables.a Note: Cylinder mass = 40Kg Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Equations of 2-D Equilibrium (1 of 2) Since particle A is in equilibrium, the net force at A is zero. So FB + FC + FD = 0 or F =0 In general, for a particle in equilibrium,  F = 0 or F i + F X y j = 0 = 0i +0 j (a vector equation) Or, written in a scalar form, F X = 0 and  FY = 0 These are two scalar equations of equilibrium (E-of-E). They can be used to solve for up to two unknowns. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Equations of 2-D Equilibrium (2 of 2) Note: Cylinder mass = 40 Kg Write the scalar E-of-E: + →  FX = FB cos 30o − FD = 0 +   FY = FB Sin30o − 392.4N = 0 Solving the second equation gives: From the first equation, we get: Statics, Fourteenth Edition R.C. Hibbeler FB = 785N → FD = 680N  Copyright ©2016 by Pearson Education, Inc. All rights reserved. Simple Springs Spring Force = spring constant * deformation of spring or F = k*s Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Cables and Pulleys With a frictionless pulley and cable T1 = T2  Cable can support only a tension or “pulling” force, and this force always acts in the direction of the cable. Cable is in tension Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Smooth Contact If an object rests on a smooth surface, then the surface will exert a force on the object that is normal to the surface at the point of contact. In addition to this normal force N, the cylinder is also subjected to its weight W and the force T of the cord. Since these three forces are concurrent at the center of the cylinder, we can apply the equation of equilibrium to this “particle,” which is the same as applying it to the cylinder. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (1 of 2) Given: The box weighs 550 l b and geometry is as shown. Find: The forces in the ropes AB and AC. Plan: 1. Draw a FBD for point A. 2. Apply the E-of-E to solve for the forces in ropes AB and AC. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (2 of 2) Applying the scalar E-of-E at A, we get; 4 + →  FX = − FB cos30o + FC   = 0 5 3 +   F Y = FB sin30 + FC   − 550 lb = 0 5 o Solving the above equations, we get; FB = 478 lb and FC = 518 lb Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (1 of 2) Given: The mass of cylinder C is 40 kgrand geometry is as shown. Find: The tensions in cables DE, E A, and EB. Plan: 1. Draw a FBD for point E. 2. Apply the E-of-E to solve for the forces in cables D E, EA, and EB. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (2 of 2) Applying the scalar E-of-E at E, we get; + →  F X = − TED + ( 40  9.81) cos30o = 0 +   F Y = ( 40  9.81) sin30o − TEA = 0 and TE A = 196 N Statics, Fourteenth Edition R.C. Hibbeler ↓ TE D =340 N ← Solving the above equations, we get; Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (1 of 3) Given: The mass of lamp is 20 kgrand geometry is as shown. Find: The force in each cable. Plan: 1. Draw a FBD for Point D. 2. Apply E-of-E at Point D to solve for the unknowns (FCD & F D E). 3. Knowing FCD, repeat this process at point C. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (2 of 3) Applying the scalar E-of-E at D, we get; +   FY = FDE sin30o − 20 ( 9.81) = 0 + →  FX = FDE cos30o − FCD = 0 Solving the above equations, we get: FDE = 392N and FCD =340N  Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (3 of 3) Applying the scalar E-of-E at C, we get; 3 + →  FX = 340 − FBC sin 45o − FAC   = 0 5 4 +   FY = FAC   − FBC cos 45o = 0 5 Solving the above equations, we get; FBC = 275N and FAC = 243N Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. 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 Statics, Fourteenth Edition R.C. Hibbeler Copyright © 2022 Pearson Education, Inc. All Rights Reserved Copyright ©2016 by 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 counter-clockwise (C CW), depending on the tendency for rotation. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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 counterclockwise 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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). Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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., Statics, Fourteenth Edition R.C. Hibbeler i  i =0 Copyright ©2016 by 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅰ (2 of 2) 3 +  Fy = –100   N 5 4 + → Fx = 100   N 5 Solution: Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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 . Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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 Find the position vector 250 ) j + (75 + 100 ) k  lb 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 Statics, Fourteenth Edition R.C. Hibbeler = [{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 ©2016 by 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 Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by 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 D.  12  N·m  3  E. 7 N·m Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Moment About an Axis To determine the moment of a force about an axis using a. scalar analysis, and, b. vector analysis. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Scalar Analysis (1 of 2) Recall that the moment of a scalar force about any point O is MO = F dO where dO is the perpendicular (or shortest) distance distance from the point to the force’s line of action. This concept can be extended to find the moment of a force about an axis. Finding the moment of a force about an axis can help answer the types of questions we just considered. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Vector Analysis (1 of 2) Our goal is to find the moment of F (the tendency to rotate the body) about the a-axis. First compute the moment of F about any arbitrary point O that lies on the aaxis using the cross product. MO = r  F Now, find the component of M0 along the a-axis using the dot product. Ma = u a  Mo Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Vector Analysis (2 of 2) Ma can also be obtained as ua x uay uaz Ma = ua  ( r  F ) = rx Fx ry Fy rz Fz The above equation is also called the triple scalar product. In the this equation, ua represents the unit vector along the a-axis, r is the position vector from any point on the a-axis to any point A on the line of action of the force, and F is the force vector. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (1 of 2) Given: A force is applied to the tool as shown. Find: The magnitude of the moment of this force about the x axis of the value. Plan: 1. Use Mz = u • ( r  F ) . 2. First, find F in Cartesian vector form. 3. Note that u = 1 i in this case. 4. The vector r is the position vector from O to A. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (2 of 2) Solution: u =1 i rOA = 0 i + 0.3 j + 0.25 k  m F = 200 ( cos 120 i + cos 60 j + cos 45 k ) N = −100 i + 100 j + 141.4 k  N Now find Mz = u  ( rOA  F ) 1 0 0 Mz = 0 0.3 0.25 = 10.3 (141.4 ) − 0.25 (100 )N  m −100 100 141.4 Mz = 17.4N  m CCW Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (1 of 3) Given: The force of F = 30 N acts on the bracket.  = 60°,  = 60°,  = 45°. Find: The moment of F about the a-a axis. Plan: 1. Find ua and rOA 2. Find F in Cartesian vector form. 3. Use Ma = ua  ( rOA  F ) Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (2 of 3) Solution: ua = j rOA = −0.1 i + 0.15 k  m F = 30 cos 60 i + cos 60 j + cos 45 k  N F = 15 i + 15 j + 21.21k  N Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (3 of 3) Now find the triple product, Ma = ua  ( rOA  F ) 0 1 0 Ma = −0.1 0 0.15 N  m 15 15 21.21 Ma = −1−0.1( 21.21) − 0.15 (15 ) = 4.37 N  m Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Moment of a Couple a. define a couple, and, b. determine the moment of a couple. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Moment of a Couple (1 of 2) A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance “d.” The moment of a couple is defined as MO = F d (using a scalar analysis) or as MO = r  F (using a vector analysis). Here r is any position vector from the line of action of F to the line of action of F. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅰ: Scalar Approach (1 of 2) Given: Two couples act on the beam with the geometry shown. Find: The magnitude of F so that the resultant couple moment is 1.5 kN  m clockwise. Mz = u • (r  F ). Plan: 1. Add the two couples to find the resultant couple. 2. Equate the net moment to 1.5 kN  m clockwise to find F. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅰ: Scalar Approach (2 of 2) Solution: The net moment is equal to: Solving for the unknown force F, we get F = 2.33 kN Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ: Vector Approach (1 of 2) Given: A 450 N force couple acting on the pipe assembly. Find: The couple moment in Cartesian vector notation. Plan: 1. Use M = r × F to find the couple moment. 2. Set r = rAB and F = FB. 3. Calculate the cross product to find M. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ: Vector Approach (2 of 2) Solution: rAB = { 0.4 i } m 4 3 FB = {0 i + 450   j − 450   k } N 5 5 = {0 i + 360 j − 270 k } N M = rAB  FB = [{0 ( −270 ) – 0 ( 360 )}i – 4 ( −270 ) – 0 ( 0 ) j + 0.4 ( 360 ) – 0 ( 0 ) k ] N·m = {0 i +108 j + 144 k } N·m Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving I (1 of 3) Given: Two couples act on the beam with the geometry shown and d = 4 f t. Find: The resultant couple ee Plan: 1. Resolve the forces in x and ydirections so they can be treated as couples. 2. Add these two couples to find the resultant couple. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving I (2 of 3) The x and y components of the upper-left 50 lb force are: 50 lb (cos 30°) = 43.30 lb vertically up 50 lb (sin 30°) = 25 lb to the right Do both of these components form couples with their matching components of the other 50 force? No! Only the 43.30 lb components create a couple. Why? Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (3 of 3) Now resolve the lower 80 lb force: (80 lb)  3  acting up  , 5 (80 lb)  4  , acting to the right 5   Do both of these components create a couple with components of the other 80 lb force? The net moment is equal to: Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Equilibrium of a Rigid Body & Free-Body Diagrams a. Identify support reactions, and, b. Draw a free-body diagram. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 5.1: Conditions for Rigid-Body Equilibrium For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero. Forces on a particle  F = 0 ( no translation ) and  MO = 0 ( no rotation ) Forces on a rigid body Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 5.2: Free-Body Diagrams (1 of 2) Idealized model Free-body diagram (FBD) 1. Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape. 2. Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Support Reactions in 2-D A few example sets of diagrams s are shown above. Other support reactions are given in your textbook (Table 5-1). As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body in the opposite direction. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 5.2: Free-Body Diagrams (1 of 2) Idealized model Free-body diagram (FBD) 1. Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape. 2. Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 5.2: Free-Body Diagrams (2 of 2) 3. Label loads and dimensions on the FBD: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA. Indicate any necessary dimensions. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (1 of 3) Given: The beam is supported by the roller at A and a pin at B. Find: The reactions at points A and B on the beam. Plan: a. Establish the x–y axis system. b. Draw a complete FBD of the beam. c. Apply the E-of-E to solve for the unknowns. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (2 of 3) Note that the distributed load has been reduced to a single force. First, write a moment equation about point B. Why point B? Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Internal Forces (2 of 2) In two-dimensional cases, typical internal loads are normal or axial forces (N, acting perpendicular to the section), shear forces (V, acting along the surface), and the bending moment (M). The loads on the left and right sides of the section at B are equal in magnitude but opposite in direction. This is because when the two sides are reconnected, the net loads are zero at the section. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Steps for Determining Internal Forces 1. Take an imaginary cut at the place where you need to determine the internal forces. Then, decide which resulting section or piece will be easier to analyze. 2. If necessary, determine any support reactions or joint forces you need by drawing a FBD of the entire structure and solving for the unknown reactions. 3. Draw a FBD of the piece of the structure you’ve decided to analyze. Remember to show the N, V, and M loads at the “cut” surface. 4. Apply the E-of-E to the FBD (drawn in step 3) and solve for the unknown internal loads. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (1 of 3) Given: The loading on the beam. Find: The internal forces at point C. Plan: Follow the procedure!! Solution 1. Plan on taking the imaginary cut at C. It will be easier to work with the right section (from the cut at C to point B) since the geometry is simpler and there are no external loads. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (2 of 3) 2. We need to determine B y. Use a F B D of the entire frame and solve the E-of-E for B y. Applying the E-of-E to this F B D, we get → +  FX = BX = 0; Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (3 of 3) 3. Now draw a F B D of the right section. Assume directions for V C, N C and M C. 4. Applying the E-of-E to this F B D, we get → +  FX = NC = 0;  +  Fy = − VC − 6 = 0; Statics, Fourteenth Edition R.C. Hibbeler NC = 0 VC = −6 kip Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving I (1 of 3) Given: The loading on the beam. Find: The internal forces at point C. Plan: Follow the procedure!! Solution: 1. Plan on taking the imaginary cut at C. It will be easier to work with the left section (point A to the cut at C) since the geometry is simpler. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving I (2 of 3) 2. First, we need to determine A x and A y using a F B D of the entire frame. Applying the E-of-E to this F B D, we get → +  FX = A X + 400 = 0; Statics, Fourteenth Edition R.C. Hibbeler A X = −400 N Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving I (3 of 3) 3. Now draw a F B D of the left section. Assume directions for V C, N C and M C as shown. Applying the E-of-E to this F B D, we get → +  FX = NC − 400 = 0;  +  Fy = − VC − 96 = 0; Statics, Fourteenth Edition R.C. Hibbeler N C = 400 N VC = −96 N Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving II (1 of 3) Given: The loading on the beam. Find: The internal forces at point C. Plan: Follow the procedure!! Solution: 1. Make an imaginary cut at C. Why there? Which section will you pick to analyze via the F B D? Why will it be easier to work with segment A C? Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving II (2 of 3) 2. Determine the reactions at A, using a F B D and the E-of-E for the entire frame. → +  FX = A X − 4320 = 0;  +  Fy = A y − 1800 = 0; Statics, Fourteenth Edition R.C. Hibbeler A X = 4320 lb A y = 1800 lb Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving II (3 of 3) 3. A F B D of section A C is shown below. 4. Applying the E-of-E to the F B D, we get → +  FX = NC + 4320 = 0; NC = −4320 lb  +  Fy = 1800 − 450 − VC = 0; Statics, Fourteenth Edition R.C. Hibbeler VC = 1350 lb Copyright ©2016 by Pearson Education, Inc. All rights reserved. Engineering Mechanics: Statics Fourteenth Edition Chapter 8 Friction Section 8.1 Characteristics of Dry Friction Section 8.2 Problems Involving Dry Friction Section 8.3 Wedges Section 8.5 Frictional Forces on Flat Belts Copyright Statics, Fourteenth Edition R.C. Hibbeler © 2016, 2013, 2010Copyright Pearson Education, Inc. All Rights ©2016 by Pearson Education, Inc. Reserved All rights reserved. Section 8.1: Characteristics of Dry Friction Friction is defined as a force of resistance acting on a body which prevents or resists the slipping of a body relative to a second body. Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion. For the body shown in the figure to be in equilibrium, the following must be true: Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 8.1: Characteristics of Dry Friction The maximum friction force is attained just before the block begins to move (a situation that is called “impending motion”). The value of the force is found using F= 𝜇 𝑁 Where 𝑠 is called the coefficient of static friction. The value of depends on the two in contact. Once the block begins to move, the frictional force typically drops and is given by F= 𝜇 𝑁 The value of (coefficient of 𝑘 kinetic friction) is less than Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 8.1: Characteristics of Dry Friction (4 of 4) It is also very important to note that the friction force may be less than the maximum friction force. So, just because the object is not moving, don’t assume the friction force is at its maximum of F= 𝜇𝑠 𝑁 unless you are told or know motion is impending! Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Impending Tipping Versus Slipping (1 of 2) For a given W and h of the box, how can we determine if the block will slide or tip first? In this case, we have four unknowns (F, N, x, and P) and only the three E-of-E. Hence, we have to make an assumption to give us another equation (the friction equation!). Then we can solve for the unknowns using the three E-of-E. Finally, we need to check if our assumption was correct. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Impending Tipping Versus Slipping (2 of 2) Assume: Slipping occurs Known: F= 𝜇 𝑁 𝑠 Solve: x, P, and N Check: Or Assume: Tipping occurs Known: Solve: P, N, and F Check: F≤ 𝜇 𝑁 𝑠 Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Analysis of a Wedge A wedge is a simple machine in which a small force P is used to lift a large weight W. To determine the force required to push the wedge in or out, it is necessary to draw FBDs of the wedge and the object on top of it. It is easier to start with a FBD of the wedge since you know the direction of its motion. Note that: a. the friction forces are always in the direction opposite to the motion, or impending motion, of the wedge; b. the friction forces are along the contacting surfaces; and, c. the normal forces are perpendicular to the contacting surfaces. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Analysis of a Wedge Next, a FBD of the object on top of the wedge is drawn. Please note that: a. at the contacting surfaces between the wedge and the object, the forces are equal in magnitude and opposite in direction to those on the wedge; and, b. all other forces acting on the object should be shown. To determine the unknowns, we must apply E-of-E, and  Fx = 0  Fy = 0 to the wedge and the object as well as the impending motion frictional equation, Statics, Fourteenth Edition R.C. Hibbeler F = S N. Copyright ©2016 by Pearson Education, Inc. All rights reserved. Analysis of a Wedge Note: If the object is to be lowered, then the wedge needs to be pulled out. If the value of the force P needed to remove the wedge is positive, then the wedge is self-locking, i.e., it will not come out on its own. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Belt Analysis Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to  radians. If the belt slips or is just about to slip, then must be larger than T 1 T2 and the motion resisting friction must be greater forces. Hence, T2 than T1. Detailed analysis (please refer to your textbook) shows that is the coefficient of static friction between where  Be sure to use radians when using this Tthe T1 eand the surface. 2 = belt formula!!  Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (1 of 3) Given: Automobile has a mass of 2000 k g and ilo ram Find: The smallest magnitude of F required to move the car if the back brakes are locked and the front wheels are free to roll. Plan: a. Draw FBDs of the car. b. Determine the unknowns. c. Apply the E-of-E and friction equations to solve for the unknowns. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (2 of 3) Here is the correct FBD: Note that there are four unknowns: Equations of Equilibrium: Assume that the rear wheels are on the verge of slip. Thus Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving (3 of 3) Solving Equations (1) to (4), F = 2762 N and Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example 1 (1 of 3) Given: The 3000-lb load is applied to wedge B. The coefficient of static friction between A and C and between B and D is 0.3, and between A and B it is 0.4. Assume the wedges have negligible weight. Find: The smallest force P needed to lift 3000 lb load. Plan: 1. Draw FBDs of wedge A and wedge B. 2. Apply the E-of-E to wedge B. Why do wedge B first? 3. Apply the E-of-E to wedge A. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example 1 (2 of 3) Applying the E-of-E to wedge B, we get → +  FX = N sin 15 + 0.4 N cos 15 − ND = 0  +  FY = N cos 15 − 0.4 N sin 15 − 0.3 ND − 3000 = 0 Solving these two equations, we get N = 4485 lb, ND = 2894 lb Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example 1 (3 of 3) Applying the E-of-E to wedge A, we get  +  FY = NC + 0.4 (4485) sin 15 − 4485 cos 15 = 0; NC = 3868 lb → +  FX = P − 0.3 (3868) − 4485 sin 15 − 1794 cos 15 = 0; P = 4054 lb Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Engineering Mechanics: Statics Fourteenth Edition Chapter 9 Center of Gravity and Centroid Section 9.1 Center of Gravity, Center of Mass, and the Centroid of a Body Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Concept of Center of Gravity (CG) A body is composed of an infinite number of particles, and so if the body is located within a gravitational field, then each of these particles will have a weight dW. The center of gravity (CG) is a point, often shown as G, which locates the resultant weight of a system of particles or a solid body. From the definition of a resultant force, the sum of moments due to individual particle weight about any point is the same as the moment due to the resultant weight located at G. Also, note that the sum of moments due to the individual particle’s weights about point G is equal to zero. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Concept of CG The location of the center of gravity, measured from the y axis, is determined by equating the moment of W about the y-axis to the sum of the moments of the weights of the particles about this same axis. If dW is located at point ( x,y,z ) , then _ ~ x W =  x dW _ Similarly, _ ~ ~ zW =  z dW y W =  y dW Therefore, the location of the center of gravity G with respect to the x, y, z-axes becomes  x dW ~ x=  dW  y dW ~ y= Statics, Fourteenth Edition R.C. Hibbeler  dW z=  z dW  dW Copyright ©2016 by Pearson Education, Inc. All rights reserved. CM & Centroid of a Body x dW  x=  dW ~ y dW  y=  dW ~ z dW  z=  dW By replacing the W with a m in these equations, the coordinates of the center of mass can be found.  x dm ~ x=  dm  y dm ~ y=  dm z dm  z=  dm Similarly, the coordinates of the centroid of volume, area, or length can be obtained by replacing W by V, A, or L, respectively. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Concept of Centroid The centroid, C, is a point defining the geometric center of an object. The centroid coincides with the center of mass or the center of gravity only if the material of the body is homogenous (density or specific weight is constant throughout the body). If an object has an axis of symmetry, then the centroid of object lies on that axis. In some cases, the centroid may not be located on the object. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Steps to Determe the Centroid of an Area 1. Choose an appropriate differential element dA at a general point (x, y). (e.g.,y = x 2 + 1), Hint: Generally, if y is easily expressed in terms of x use a vertical rectangular element. If the converse is true, then use a horizontal rectangular element. 2. Express d A in terms of the differentiating element dx (or dy). 3. Determine coordinates ~ ~ ( x, y ) of the centroid of the rectangular element in terms of the general point (x, y). 4. Express all the variables and integral limits in the formula using either x or y depending on whether the differential element is in terms of dx or dy, respectively, and integrate. Note: Similar steps are used for determining the CG or CM. These steps will become clearer by doing a few examples. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (1 of 2) Given: The area as shown. Find: The centroid location Plan: Follow the steps. (x,y) Solution: 1. Since y is given in terms of x, choose dA as a vertical rectangular strip. 2. 3. dA = y dx = x 3 dx ~ x=x y x3 y= = 2 2 ~ and Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (2 of 2) 4. x = (  A x dA ( A dA ) ) 1 ( ) 3 x x dx  = 0 1 (x )d x 3 0 1 5 1  x  0 = 5 1 4 1 x  4  0  1 5 =   =0.8 m  1 4    X3  3 1 7 1  x  A y dA 0  2  X dx 14 0 y= = = 1 1 3 A dA X dx  4 1 ( ) 0  1  14  =   =0.2857 m  1 4   Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example II Given: The shape and associated horizontal rectangular strip shown. Find: dA and ( x,y ) Plan: Follow the steps. Solution: 1. dA = x dy = y2dy 2. 1 - x ) (1 ( x=x+ = ~ 2 3. + x) 2 1+ y ) ( = 2 2 % y= y Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example III Locate the centroid y of the area Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Locate the centroid y of the area Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example IV Determine the area and the centroid x of the parabolic area. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Reading Quiz (1 of 2) 1. The Fill in the blank is the point defining the geometric center of an object. A. Center of gravity B. Center of mass C. Centroid D. None of the above Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Attention Quiz (1 of 2) 1. If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of Fill in the blank . A. x B. y C. z D. Any of the above. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Definition of Moments of Inertia for Areas, Radius of Gyration of an Area Today’s Objectives: a. Define the moments of inertia (MoI) for an area. b. Determine the MoI for an area by integration. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 10.1: Definition of Moments of Inertia for Areas For the differential area dA shown in the figure: d Ix = y 2 dA, d Iy = x 2 dA, and, d JO = r 2 dA, where JO is the polar moment of inertia about the pole O or z axis. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Section 10.1: Definition of Moments of Inertia for Areas The moments of inertia for the entire area are obtained by integration. Ix =  y 2 dA ; Iy = A  A x 2 dA JO =  r 2 dA =  (x 2 + y 2 )dA = Ix + Iy A A The MoI is also referred to as the second moment of an area and has units of length to the fourth power (m4 or in4 ). Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. MoI for an Area by Integration (1 of 3) For simplicity, the area element used has a differential size in only one direction (dx or dy). This results in a single Integration and is usually simpler than doing a double integration with two differentials, i.e., dx  dy. The step-by-step procedure is: 1. Choose the element dA : There are two choices: a vertical strip or a horizontal strip. Some considerations about this choice are: Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. MoI for an Area by Integration (2 of 3) a. The element parallel to the axis about which the MoI is to be determined usually results in an easier solution. For example, we typically choose a horizontal strip for determining and a vertical strip for determining Ix Iy . b. If y is easily expressed in terms of x (e.g., y = x 2 + 1), then choosing a vertical strip with a differential element dx wide may be advantageous. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. MoI for an Area by Integration (3 of 3) 2. Integrate to find the MoI. For example, given the element shown in the figure above: Iy =  x 2 dA =  x 2 y dx (using the parallel axis theorem as  1 Ix =  dIx =    y 3 dx 3 per Example 10.2 of the textbook). Since the differential element is dx, y needs to be expressed in terms of x and the integral limit must also be in terms of x. As you can see, choosing the element and integrating can be challenging. It may require a trial and error approach, plus experience. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (1 of 2) Given: The shaded area shown in the figure. Find: The MoI of the area about the xand y-axes. Plan: Follow the steps given earlier. Solution: Ix = y 2 dA 3 2 dA = (1 − x)dy = (1 − y )dy 1 Ix = y 2 3 2 (1 − y )dy 0 1  1  3  2  92  =   y −   y  = 0.111m4  9  0  3  Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example (2 of 2) Iy = x = = 2 dA = x 2 2 3 1 8 3 x x 2 y dx (x ) dx dx 0 1  3  11  3 =  x  11   0 = 0.273 m4 In the above example, Then Ix can be also determined using a vertical strip.  1 Ix =    y 3 dx = 3 Statics, Fourteenth Edition R.C. Hibbeler 1 1  1 2 4 x dx = = 0.111 m 0  3  9 Copyright ©2016 by Pearson Education, Inc. All rights reserved. Concept Quiz (2 of 2) 2. In the figure to the right, what is the differential moment of inertia of the element with respect to the y-axis A. x 2 y dx B.  1 3   x dy  12  C. y 2 x dy D.  1   y dy 3 Statics, Fourteenth Edition R.C. Hibbeler (dIy )? Copyright ©2016 by Pearson Education, Inc. All rights reserved. Determine the moment of inertia for the shaded area about the x axis. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Determine the moment of inertia for the shaded area about the x axis. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved.

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