ME 221Lecture 51 ME 221 Statics LECTURE #4 Sections:

ME 221Lecture 52 Announcements HW #2 due Friday 5/28 Ch 2: 23, 29, 32, 37, 47, 50, 61, 82, 105, 113 Ch 3: 1, 8, 11, 25, 35 Quiz #3 on Friday, 5/28 Exam #1 on Wednesday, June 2

ME 221Lecture 53 Chapter 3 Rigid Bodies; Moments Consider rigid bodies rather than particles –Necessary to properly model problems Moment of a force about a point Problems Moment of a force about an axis Moment of a couple Equivalent force couple systems

ME 221Lecture 54 Rigid Bodies The point of application of a force is very important in how the object responds F F We must represent true geometry in a FBD and apply forces where they act.

ME 221Lecture 55 Transmissibility A force can be replaced by an equal magnitude force provided it has the same line of action and does not disturb equilibrium B A

ME 221Lecture 56 Moment A force acting at a distance is a moment Transmissibility tells us the moment is the same about O or A F d M O M A d is the perpendicular distance from F’s line of action to O Defn. of moment: M = F d

ME 221Lecture 57 Vector Product; Moment of Force Define vector cross product –trig definition –component definition cross product of base vectors Moment in terms of cross product

ME 221Lecture 58 Cross Product The cross product of two vectors results in a vector perpendicular to both. B A  A x B The right-hand rule decides the direction of the vector. B x A A B  A x B = - B x A n = AxB ^

ME 221Lecture 59 Base Vector Cross Product Base vector cross products give us a means for evaluating the cross product in components. Here is how to remember all of this: +-

ME 221Lecture 510 General Component Cross Product Consider the cross product of two vectors ˆ A z B y  i Or, matrix determinate gives a convenient calculation

ME 221Lecture = (A y B z -A z B y ) i - (A x B z -A z B x ) j + (A x B y -A y B x )k

ME 221Lecture 512 Example Problems If: A = 5i + 3j & B = 3i + 6j Determine: A·B The angle between A and B AxB BxA

ME 221Lecture 513

ME 221Lecture 514 Vector Moment Definition The moment about point O of a force acting at point A is: M O = r A/O x F Compute the cross product with whichever method you prefer. r A/O A O F

ME 221Lecture 515 A O N 60 o x d 0.285tan 60°=0.2m/x x=0.115m sin 60°=d/0.285m d = m M A =200N *0.247m= 49.4 Nm Example Method # 1

ME 221Lecture 516 A O N 60 o 200 cos sin 60 M+ =200N (sin 60)(0.4m)- 200N (cos 60)(0.2m) = 49.4 Nm Method # 2 Note: Right-hand rule applies to moments

ME 221Lecture 517 A O N 60 o Method # 3 r F=200N cos 60 i + 200N sin 60 j r =0.4 i j cos60 200sin60 0 MA=MA= =200 (sin 60)(0.4) (cos 60)(0.2) = 49.4 Nm i j k ^ ^ ^

ME 221Lecture 518 A O N 60 o Method # 4 r =0.285 i F=200N cos 60 i + 200N sin 60 j r =0.285 i i j k cos60 200sin60 0 MA=MA= = 49.4 Nm

ME 221Lecture 519 Moment of a Force about an Axis x y z O F B n ^ A r AB =r B/A |M n | =M A ·n ^ =n·(r B/A x F ) ^ Same as the projection of M A along n | M n |= n x n y n z r B/Ax r B/A y r B/Az F x F y F z

ME 221Lecture 520 x y z O F B n ^ A r AB =r B/A MAMA MnMn Mp Resolve the vector M A into two vectors one parallel and one perpendicular to n. M n =|M n |n ^ M p = M A - M n =n x [(r B/A x F) x n] ^

ME 221Lecture 521 Moment of a Couple x y z O F2F2 B rArA F1F1 r AB =r B/A rBrB A Let F 1 = -F 2 d M o =r A x F 2 + r B x F 1 =(r B - r A ) x F 1 =r AB x F 1 = C The Moment of two equal and opposite forces is called a couple |C|=|F 1 | d

ME 221Lecture 522 The two equal and opposite forces form a couple (no net force, pure moment) The moment depends only on the relative positions of the two forces and not on their position with respect to the origin of coordinates Moment of a Couple (continued)

ME 221Lecture 523 Since the moment is independent of the origin, it can be treated as a free vector, meaning that it is the same at any point in space The two parallel forces define a plane, and the moment of the couple is perpendicular to that plane Moment of a Couple (continued)

ME 221Lecture 524 Example