Section 3.1 Condition for the Equilibrium of a Particle

Engineering Mechanics: Statics Fifteenth Edition Chapter 3 Equilibrium of a Particle Section 3.1 Condition for the Equilibrium of a Particle Section 3.2 The Free-Body Diagram Section 3.3 Coplanar Force Systems Statics, Fourteenth Edition R.C. Hibbeler Copyright © 2022 Pearson Education, Inc. All Rights Reserved 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. Attention Quiz (2 of 2) 2. Using this FBD of Point C, the sum of forces in the x-direction (F ) X Is ______. Use a sign convention of + → . A. F2 sin50o − 20 = 0 B. F2 cos50o − 20 = 0 C. F2 sin50o − F1 = 0 D. F2 cos50o + 20 = 0 Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. The Equations of 3-D Equilibrium (1 of 2) When a particle is in equilibrium, the vector sum of all the forces acting on it must be zero ( F = 0 ). This equation can be written in terms of its x, y and z components. This form is written as follows. ( Fx ) i + ( Fy ) j + ( Fz ) k = 0 Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. The Equations of 3-D Equilibrium (2 of 2) This vector equation will be satisfied only when Fx = 0 Fy = 0 Fz = 0 These equations are the three scalar equations of equilibrium. They are valid for any point in equilibrium and allow you to solve for up to three unknowns. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (1 of 3) Given: The four forces and geometry shown. Find: The tension developed in cables A B, A C, and A D. Plan: 1. Draw a FBD of particle A. 2. Write the unknown cable forces TB, TC , and TD in Cartesian vector form. 3. Apply the three equilibrium equations to solve for the tension in cables. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (2 of 3) Solution: TB = TB i TC = −(TC cos 60°) sin30°i + (TC cos 60°) cos30° j +TC sin60°k TC = TC(−0.25i +0.433 j + 0.866k ) TD = TD cos 120° i + TD cos120° j + TD cos45° k TD = TD(−0.5 i − 0.5 j + 0.7071 k ) W = − 300 k Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅰ (3 of 3) Applying equilibrium equations: FR = 0 = TB i + TC(− 0.25 i +0.433 j + 0.866 k ) + TD(−0.5 i − 0.5 j + 0.7071k ) − 300k Equating the respective i, j, k components to zero, we have Fx = TB – 0.25TC –0.5TD =0 Fy = 0.433 TC –0.5TD =0 Fz = 0.866TC + 0.7071TD – 300 =0 (1 ) ( 2) (3) Using (2) and (3), we can determine Substituting TC and TD into (1), we can find Statics, Fourteenth Edition R.C. Hibbeler TC = 203 lb , TD = 176 lb TB = 139 lb Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (1 of 3) Given: A 600 N load is supported by three cords with the geometry as shown. Find: The tension in cords AB, Aand A D. Plan: 1. Draw a free body diagram of Point A. Let the unknown force magnitudes be FB, FC, FD. 2. Represent each force in its Cartesian vector form. 3. Apply equilibrium equations to solve for the three unknowns. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (2 of 3) FB = FB(sin 30 i + cos30 j ) N F = F  rAD  D D   rAD  = {0.5 FB i +0.866 FB j } N  FC = –FC i N = FD  (1 i – 2  j + 2k) N 1   (12 + 2 2 + 2) 2    = { 0.333 FD i – 0.667 FD j + 0.667 FD k } N Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (3 of 3) Now equate the respective i, j, and k components to zero.  Fx = 0.5 FB – FC + 0.333 FD = 0  Fy = 0.866 FB – 0.667 FD = 0  Fz = 0.667 FD – 600 = 0 Solving the three simultaneous equations yields FC = 646 N (since it is positive, it is as assumed, e.g., in tension) FD = 900 N FB = 693 N 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. 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. Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Group Problem Solving Ⅰ (2 of 2) Solution: +  Fy = 20 sin 30° lb + → Fx = 20 cos 30° lb 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 (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. Engineering Mechanics: Statics Fifteenth Edition Chapter 4 Force System Resultants Section 4.7 Simplification of a Force and Couple System Section 4.8 Further Simplification of a Force and Couple System Statics, Fourteenth Edition R.C. Hibbeler Copyright © 2022 Pearson Education, Inc. All Rights Reserved Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example I (1 of 3) Given: A 2-D force system with geometry as shown. Find: The equivalent resultant force and couple moment acting at A and then the equivalent single force location measured from A. Plan: 1. Sum all the x and y components of the forces to find FRA. 2. Find and sum all the moments resulting from moving each force component to A. 3. Shift FRA to a distance d such that Statics, Fourteenth Edition R.C. Hibbeler d= MRA FRy Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅰ (2 of 3) 3 + →  FRx = 50 ( sin 30 ) + 100   5 = 85lb 4 +   FRy = 200 + 50 ( cos 30 ) − 100   5 = 163.3lb Statics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅰ (3 of 3) FR = ( 852 + 163.3 1 2 2 ) = 184 lb The equivalent single force FR can be located at a distance d measured from A. D = 𝑀𝑅𝐴 = 𝐹𝑅 Statics, Fourteenth Edition R.C. Hibbeler 509 7 . 184 = 2.77 𝑓𝑡 Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (1 of 3) Given: The slab is subjected to three parallel forces. Find: The equivalent resultant force and couple moment at the origin O. Also find the location (x, y) of the single equivalent resultant force. Plan: 1. Find FRO =  Fi = FRzo k 2. Find MRO =  ( ri  Fi ) = MRXO i + MRYO j 3. The location of the single equivalent resultant force is given as x=− MRyO FRzO and y = Statics, Fourteenth Edition R.C. Hibbeler MRxO FRzO Copyright ©2016 by Pearson Education, Inc. All rights reserved. Example Ⅱ (2 of 2) FRO = {100 k − 500 k − 400 k } = − 800 k N MRO = (3 i )  (100 k ) + (4 i + 4 j )  ( −500 k ) + (4 j )  ( −400 k ) = {−300 j + 2000 j − 2000 i – 1600 i } = { – 3600 i + 1700 j } N  m The location of the single equivalent resultant force is given as, ( −1700 ) = 2.13 m FRzo ( −800 ) ( −3600 ) = 4.5 m M y = Rxo = FRzo ( −800 ) x=− MRyo Statics, Fourteenth Edition R.C. Hibbeler = Copyright ©2016 by Pearson Education, Inc. All rights reserved.