ME 201 Engineering Mechanics: Statics Chapter 4 – Part A 4.1 Moment of a Force - Scalar 4.2 Cross Product 4.3 Moment of a Force – Vector 4.4 Principle of Moments

Moment of a Force Scalar Formulation  The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.  Example: wrench

Moment of a Force Scalar Formulation M O = F d Where d = moment arm or perpendicular distance from axis O to the line of action of the force Line of action d F O

Moment of a Force Scalar Formulation  What if multiple forces? Multiply each force by its respective length and sum  Typically easier to break a force into horizontal and vertical components in order to compute associated moment

Moment of a Force Scalar Formulation  Typically easier to break a force into horizontal and vertical components in order to compute associated moment d F O FxFx O FyFy d1d1 d2d2

Example Problem Given: F 1 = 50 N F 2 = 100 N θ = 30º Find: M O θO 3 m 1 m F1F1 F2F2

Example Problem Solution Given: F 1 = 50 N F 2 = 100 N θ = 30º Find: M O Solution: θO 3 m 1 m F1F1 F2F2 F 2y F 1x d F2 d F1 What about F 2x ? F 2x pulls on O but does not cause rotation

Cross Product  Cross Product is a vector operation used in computing moments in 3-dimensions  The result of a cross product operation is a vector C = A x B  Magnitude of C C = A B sin θ  Direction Perpendicular to the plane containing A and B such that the direction of C is specified by right-hand rule

Cross Product  Stated another way: C = A x B = A B sin θ U C θ A B UCUC C=AB sin θ C

Cross Product Laws of Operation  Not Commutative A x B ≠ B x A However, A x B = -B x A  Multiplication by Scalar a(A x B) = (aA) x B = A x (aB) = (A x B)a  Distributive Law A x (B + D) = (A x B) + (A x D)

Cross Product Cartesian Vector Formulation  Sign convention of cross products follows right hand rule: i x j = k i * j * sin 90 = 1*1*1 = 1 i x k = -j i * k * sin 90 = 1*1*1 = 1 i x i = 0 i * i * sin 0 = 1*1*0 = 0 i j k z yx

Cross Product Cartesian Vector Formulation  Alternately, sign of the cross product of unit vectors may be determined from the following diagram. CCW direction yields + i x j = k CW direction yields – i x k = -j i j k + -

Cross Product Cartesian Vector Formulation AxB = (A x i +A y j + A z k) x (B x i + B y j + B z k) =A x B x (i x i) + A x B y (i x j) +A x B z (i x k) +A y B x (j x i) + A y B y (j x j) +A y B z (j x k) +A z B x (k x i) + A z B y (k x j) + A z B z (k x k) =A x B y k + A x B z (-j) + A y B x (-k) +A y B z i +A z B x j + A z B y (-i) =(A y B z -A z B y )i +(A z B x -A x B z )j + (A x B y -A y B x )k k -j -ki j -i

Cross Product Cartesian Vector Formulation Or Taking minors (with j negative) =(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

Moment of a Force Vector Formulation M O = r x F where r = position vector from O to any point lying on the line of action of F (Principle of Transmissibility)  Magnitude of M M = r F sin θ = F(r sin θ) = F d  Direction Perpendicular to the plane containing r and F such that the direction of M is specified by right-hand rule  Vector Notation M O =(r y F z -r z F y )i – (r x F z -r z F x )j +(r x F y -r y F x )k

Example Problem Given: F = 60 N Find: M A z x y 2 m F 1 m 3 m 4 m 3 m A B C

Example Problem Solution Given: F = 60 N Find: MAMA Solution: 1-Identify Coordinates 2-Formulate Vectors z x y 2 m F 1 m 3 m 4 m 3 m A B C or (0,0,0) (1,3,2) (3,4,0)

Solution: 2-Formulate Vectors z x y 2 m F 1 m 3 m 4 m 3 m A B C (0,0,0) (1,3,2) (3,4,0) Example Problem Solution

Solution: 2-Formulate Vectors z x y 2 m F 1 m 3 m 4 m 3 m A B C or Example Problem Solution (0,0,0) (1,3,2) (3,4,0)

Solution: 3-Perform Cross Product z x y 2 m F 1 m 3 m 4 m 3 m A B C Example Problem Solution (0,0,0) (1,3,2) (3,4,0)

Solution: 3-Perform Cross Product z x y 2 m F 1 m 3 m 4 m 3 m A B C Magnitude and direction angles may also be found from the moment vector, as required: Example Problem Solution (0,0,0) (1,3,2) (3,4,0)

Solution: 3-Perform Cross Product z x y 2 m F 1 m 3 m 4 m 3 m A B C Using alternate position vector produces same results: Example Problem Solution (0,0,0) (1,3,2) (3,4,0)