EML 3004 C Chapter 3 Force System Resultants

EML 3004 C Chapter 3: Force System Resultants Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -1

EML 3004 C Cross Product The Cross product of two vectors Magnitude: C=AB sin Direction: C is perpendicular to both A and B Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -2

EML 3004 C Laws of Operation for Cross Product Commutative law is not valid Scalar Multiplication Distributive Law Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -3

EML 3004 C Cartesian Vector Formulation Namas Chandra Introduction to Mechanical engineering (sec 3. 1) Hibbeler Chapter 3 -4

EML 3004 C Cross Product of Two Vectors (sec 3. 1) Let Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -5

EML 3004 C Moment Systems (sec 3. 2) The moment of a force about an axis (sometimes represented as a point in a body) is the measure of the force’s tendency to rotate the body about the axis (or point). The magnitude of the moment is: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -6

EML 3004 C Moment Systems of System of Forces (sec 3. 2) It is customary to assume CCW as the positive direction. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -7

EML 3004 C Moment of a Force-Vector Formulation Namas Chandra Introduction to Mechanical engineering (sec 3. 3) Hibbeler Chapter 3 -8

EML 3004 C Moment of a Force-Vector Formulation (sec 3. 3) The axis of the moment is perpendicular to the plane that contains both F and r The axis passes through point O Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -9

EML 3004 C Moment of Force Systems-Vector Formulation (sec 3. 3) Let a system of forces act upon a body. We like to compute the net moment of all the forces about the point O. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -10

EML 3004 C Problem 3 -10 (page 84, Section 3. 1 -3. 3) 3. 10 Determine the resultant moment about point B on the three forces acting on the beam. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -11

EML 3004 C Problem 3 -20 (page 86, Section 3. 1 -3. 3) 3. 20 The cable exerts a 140 -N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i. e. , by using a position vector from A to C, then A to B. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -12

EML 3004 C 3. 20 The cable exerts a 140 -N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i. e. , by using a position vector from A to C, then A to B. Solution-Con’t Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -13

EML 3004 C 3. 20 The cable exerts a 140 -N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i. e. , by using a position vector from A to C, then A to B. Solution-Con’t Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -14

EML 3004 C Section 3. 1 -3. 3 (In-class Exercise) Solution: Net moment should be zero Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -15

EML 3004 C Section 3. 1 -3. 3 (In-class Challenge Exercise) Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -16

EML 3004 C 3. 4 Principle of Moments (sec 3. 4) The moment of a force is equal to the sum of the moment of the force’s component about a point. (Varginon’s theorem 1654 -1722) Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -17

EML 3004 C 3. 5 Moment of a Force about a specified axis If we need the moment about other axis still through O, we can use either scalar or vector analysis. Here we have F=20 N applied. Though the typical equation gives moment with respect to b-axis, we require it through y-axis. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -18

EML 3004 C 3. 5 Moment of a Force about a specified axis-2 Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -19

EML 3004 C 3. 5 Moment of a Force about a specified axis-3 The two steps in the previous analysis can be combined with the definition of a scalar triple product. Since dot product is commutative Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -20

EML 3004 C 3. 6 Moment of a Couple-1 A couple is defined as two parallel forces with same magnitude and opposite direction. Net force is zero, but rotates in specified direction. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -21

EML 3004 C 3. 6 Equivalent Couples -2 Two couples are equivalent if they produce the same moment. The forces should be in the same or parallel planes for two couple to be equivalent. Couple moments are free vectors. They can be added at any point P in the body. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -22

EML 3004 C Problem 3 -39 (page 95, Section 3. 4 -3. 6) 3. 39 The bracket is acted upon by a 600 -N force at A. Determine the moment of this force about the y axis. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -23

EML 3004 C 3. 39 The bracket is acted upon by a 600 -N force at A. Determine the moment of this force about the y axis. Solution-Con’t Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -24

EML 3004 C Problem 3 -54 (page 103, Section 3. 4 -3. 6) 3. 54 Two couples act on the frame. If d = 6 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 3 -14) and (b) summing the moments of all the force components about point A Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -25

EML 3004 C 3. 54 Two couples act on the frame. If d = 6 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 3 -14) and (b) summing the moments of all the force components about point A Solution-Con’t Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -26

EML 3004 C Section 3. 4 -3. 6 (In-class Exercise) Two couples act on the frame. If d = 4 ft, find the resultant couple moment by (a) direct method, and (b) resolving the x and y components (take moment about A). Solution: a. Find the normal distance for each case first. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -27

EML 3004 C Section 3. 4 -3. 6 (In-class Challenge Exercise) The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -28

EML 3004 C Section 3. 4 -3. 6 (In-class Challenge Exercise-2) Solution—contd. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -29

EML 3004 C 3. 7 Movement of a Force on a Rigid Body-1 A single force on a body can cause it to rotate (moment) and translate (force). In the first example, the ruler causes a force F and in addition a moment M=Fd. In the example, the ruler causes a force F and NO ADDITIONAL moment. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -30

EML 3004 C 3. 7 Movement of a Force on a Rigid Body-2 Extend this idea to a general 3 -D case. Now, the force can be moved Force now causes the force at any point 0 and then a couple. Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -31

EML 3004 C 3. 8 Resultant of a Force and Couple System-1 Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -32

EML 3004 C 3. 9 Further Reduction on Force/Couples-1 Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -33

EML 3004 C 3. 9 Further Reduction on Force/Couples-2 Concurrent Force Systems Only equivalent force Coplanar Force Systems Namas Chandra Introduction to Mechanical engineering A single force at d from point 0 Hibbeler Chapter 3 -34

EML 3004 C 3. 9 Further Reduction on Force/Couples-2 Parallel Force Systems Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -35

EML 3004 C Problem 3 -103 (page 124, Section 3. 7 -3. 9) 3. 103 The weights of the various components of the truck shown. Replace this system of forces by an equivalent resultant force and couple moment acting at point A. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -36

EML 3004 C Problem 3 -93 (page 122, Section 3. 7 -3. 9) 3. 93 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -37

EML 3004 C Problem 3 -69 (page 119, Section 3. 7 -3. 9) 3. 69 The gear is subjected to the two forces shown. Replace these forces by an equivalent resultant force and couple moment acting at point O. Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -38

EML 3004 C 3. 69 The gear is subjected to the two forces shown. Replace these forces by an equivalent resultant force and couple moment acting at point O. Solution-Con’t Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -39

EML 3004 C Section 3. 7 -3. 9 (In-class Exercise) Solution: Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -40

EML 3004 C Section 3. 7 -3. 9 (In-class Exercise. . 2) Solution (contd). Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -41

EML 3004 C Section 3. 7 -3. 9 (In-class Challenge Exercise) The weights of the various components of the truck are shown. Replace this system by an equivalent resultant force and specify its location from point A. Solution: Equivalent force Location of force Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -42

EML 3004 C Chapter 3: Force System Resultants…concludes Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3 -43