Moment of a forceMoment of a force

The moment of a force about a point or axis provides The moment of a force about a point or axis provides a measure of the tendency of the force to cause a a measure of the tendency of the force to cause a body to rotate about the point or axisbody to rotate about the point or axis

Sometimes called “torque” but most often called Sometimes called “torque” but most often called “moment of a force” or simply the moment“moment of a force” or simply the moment

Determining the moment of a force Determining the moment of a force (2-D scalar form)(2-D scalar form)

The magnitude of the moment MThe magnitude of the moment MOO of a force F about a point O is of a force F about a point O is F d, where d is the perpendicular distance from O to the line of F d, where d is the perpendicular distance from O to the line of action of the force F action of the force F SHOWSHOW

The larger the force F, or the greater the distance d, the more The larger the force F, or the greater the distance d, the more pronounced the effectpronounced the effect

The direction of MThe direction of MOO is specified using the “right-hand rule” – the is specified using the “right-hand rule” – the fingers of the right hand are curled such that they follow the fingers of the right hand are curled such that they follow the sense of rotation (if the force could rotate about the point O), sense of rotation (if the force could rotate about the point O), the thumb then points along the moment axisthe thumb then points along the moment axis

The moment of a force will be positive if it is directed along the The moment of a force will be positive if it is directed along the +z axis and negative if it is directed along the –z axis+z axis and negative if it is directed along the –z axis

If the line of action of the force, F, passes through the point, O, If the line of action of the force, F, passes through the point, O, d=0 – no moment produced about O – no tendency for rotation d=0 – no moment produced about O – no tendency for rotation possiblepossible

The dimensions of moments are “distance x force” (i.e. N-m or The dimensions of moments are “distance x force” (i.e. N-m or lb-ft)lb-ft)

Can determine the sum of the moments of a system of forces Can determine the sum of the moments of a system of forces about a point if the forces are 2-D (coplanar) and the point lies in about a point if the forces are 2-D (coplanar) and the point lies in the same plane (can add all the moment vectors algebraically the same plane (can add all the moment vectors algebraically since all the moment vectors are collinear)since all the moment vectors are collinear)

Cross productCross product

Cross product, Cross product, AA x x BB, “, “AA cross cross BB”” Result of cross product is a vector (vector product of vectors)Result of cross product is a vector (vector product of vectors) C C == A A x x BB = (AB sin = (AB sin θθ) ) uucc

– Magnitude: AB sin Magnitude: AB sin θθ– Direction: perpendicular to the plane containing Direction: perpendicular to the plane containing AA and and BB and is and is

specified by the right-hand rule (curling the fingers of the right specified by the right-hand rule (curling the fingers of the right hand from hand from AA (cross) to (cross) to BB, the thumb points in the direction of the , the thumb points in the direction of the uucc))

Laws of operationLaws of operation– Commutative law is not valid: Commutative law is not valid: AA x x BB ≠ ≠ BB x x A, AA, A x x BB = - = -B B x x AA– Multiplication by a scalar: a (Multiplication by a scalar: a (AA x x BB) = (a ) = (a AA) x ) x BB = = AA x (a x (a BB) = () = (AA

x x BB) a) a– Distributive law: Distributive law: AA x ( x (BB + + DD) = () = (AA x x BB) + () + (AA x x DD))

Cartesian vector formulationCartesian vector formulation

i x i = (1)(1) sin 0° = (1)(1)(0) = 0i x i = (1)(1) sin 0° = (1)(1)(0) = 0 i x j = (1)(1) sin 90° = (1)(1)(1), direction by right-hand i x j = (1)(1) sin 90° = (1)(1)(1), direction by right-hand

rule, k (rule, k (SHOWSHOW)) i x k = (1)(1) sin 90° = (1)(1)(1), direction by right-hand i x k = (1)(1) sin 90° = (1)(1)(1), direction by right-hand

rule, -jrule, -j Similarly, Similarly, i x i = 0i x i = 0 j x i = -kj x i = -k k x i = jk x i = j

i x j = ki x j = k j x j = 0j x j = 0 k x j = -ik x j = -i

i x k = -ji x k = -j j x k = ij x k = i k x k = 0k x k = 0

Cross product of two general Cross product of two general vectorsvectors

AA x x BB = (A = (Axxi + Ai + Ayyj + Aj + AZZk) x (Bk) x (Bxxi + Bi + Byyj + Bj + BZZk)k)

= A= AxxBBxx (i x i) + A (i x i) + AxxBByy (i x j) + A (i x j) + AxxBBZZ (i x k) (i x k)

+ A+ AyyBBxx (j x i) + A (j x i) + AyyBByy (j x j) + A (j x j) + AyyBBzz (j x k) (j x k)

+ A+ AzzBBxx (k x i) + A (k x i) + AzzBByy (k x j) + A (k x j) + AzzBBzz (k x k) (k x k)

= A= AxxBBxx (0) + A (0) + AxxBByy (k) + A (k) + AxxBBZZ (-j) (-j)

+ A+ AyyBBxx (-k) + A (-k) + AyyBByy (0) + A (0) + AyyBBzz (i) (i)

+ A+ AzzBBxx (j) + A (j) + AzzBByy (-i) + A (-i) + AzzBBzz (0) (0)

= (A= (AyyBBzz – A – AzzBByy)i + (A)i + (AzzBBxx – A – AxxBBZZ)j + (A)j + (AxxBByy – A – AyyBBxx)k, or)k, or

= (A= (AyyBBzz – A – AzzBByy)i - (A)i - (AxxBBZ Z - A- AzzBBxx)j + (A)j + (AxxBByy – A – AyyBBxx)k)k

Determinant form, Determinant form, SHOWSHOW

Moment of a force (vector form)

Mo = r x F, where r is the position vector from O to any point on the line of action of F

Magnitude of MO: MO = r F sin θ = (r sin θ) F = d F SHOW Direction of MO follows the right-hand rule as it applies to

the cross product, “r cross F” – extending r along its line of action, the fingers of right hand are curled such that they follow the sense of rotation – the thumb then points along the moment axis – giving the direction of the moment vector

Result does not depend on where r intersects the line of action of F, rOB x F = (rOA + rAB) x F = rOA x F SHOW

Principle of transmissibility: F is a sliding vector – it can act at any point along its line of action and still create the same moment about point O

A moment is denoted with a circular arrow around the vector (a curl) SHOW

In Cartesian vector form

MO = r x F =

MO = (ryFz – rzFy) i - (rxFz – rzFx) j + (rxFy – ryFx) k

++ -- ++

ii jj kk

rrxx rryy rrzz

FFxx FFyy FFzz

Principle of Moments(Varignon’s Theorem)

The moment of a force about a point is equal to the sum of the moments of the force’s components about the point (SHOW)MO = (r x F1) + (r x F2) = r x (F1 + F2)

EXAMPLES (pg 133 - 138)

Moment of a force about a lineMoment of a force about a line(scalar analysis)(scalar analysis)

Two-step processTwo-step process Direct process - if the line of action of the force Direct process - if the line of action of the force FF is is

perpendicular to any specified axis aa, the magnitude of the perpendicular to any specified axis aa, the magnitude of the moment of moment of FF about the axis can be determined from M about the axis can be determined from Maa = F = F ddaa, where d, where daa is the perpendicular or shortest distance from is the perpendicular or shortest distance from the force line of action to the axisthe force line of action to the axis

Moment of a force about a lineMoment of a force about a line(vector analysis)(vector analysis)

Component of MO parallel to aa', Ma = ua ∙ MO, where ua is a unit vector along aa’

Substituting MO = r x F in above, Ma = ua ∙ (r x F) Can write MCan write Maa = = ua ∙ MO = = ua ∙ (r x F) as a triple scalar productas a triple scalar product

MMaa = = ua ∙ MO =

The scalar MMaa determines both the magnitude and sense of direction of Ma. If MMaa is positive, Ma will have the same sense as ua, and if MMaa is negative, then Ma will act opposite to ua.

Expressing Ma as a Cartesian vector, Ma = MMaa ua = [ua ∙ (r x F)] ua

EXAMPLES (pgs 145 - 147)EXAMPLES (pgs 145 - 147)

++ -- ++

uuaxax uuayay uuazaz

rrxx rryy rrzz

FFxx FFyy FFzz

A coupleA couple

A couple consists of two forces that have equal A couple consists of two forces that have equal magnitudes, same directions, but opposite sensesmagnitudes, same directions, but opposite senses

Parallel, distinct lines of action Parallel, distinct lines of action SHOWSHOW The resultant force is zero, the only effect is to The resultant force is zero, the only effect is to

produce a rotationproduce a rotation

Moment of a couple (scalar Moment of a couple (scalar formulation)formulation)

The moment of the couple The moment of the couple MM is defined as having a is defined as having a magnitude of M = F dmagnitude of M = F dF is the magnitude of one of the forcesF is the magnitude of one of the forces

d is the perpendicular distance or moment arm between d is the perpendicular distance or moment arm between the forcesthe forces

The direction and sense of the couple moment are The direction and sense of the couple moment are determined by the right-hand rule (where the thumb determined by the right-hand rule (where the thumb indicates the direction when the fingers are curled indicates the direction when the fingers are curled with the sense of rotation caused by the two forces – with the sense of rotation caused by the two forces – MM acts perpendicular to the plane containing the two acts perpendicular to the plane containing the two forces)forces)

Moment of a couple (vector Moment of a couple (vector formulation)formulation)

A couple moment: is equal to the sum of the moments A couple moment: is equal to the sum of the moments of both couple forces determined about any arbitrary of both couple forces determined about any arbitrary point O in space point O in space SHOWSHOWMM = = rrAA x (- x (-FF) + ) + rrBB x x FF

MM = - = -rrAA x x FF + + rrBB x x FF

MM = ( = (rrBB + - + -rrAA) x ) x FF = ( = (rrBB – – rrAA) x ) x FF = = rr x x FF

A couple moment is a free vector – it can act at any A couple moment is a free vector – it can act at any point, since point, since MM only depends on only depends on rr, not , not rrAA or or rrBB

r is directed from any point on the line of action of one of the forces to any point on the line of action of the other force

EXAMPLES (pg 155 - 159)EXAMPLES (pg 155 - 159)

Equivalent systemsEquivalent systems

A system of forces and couple moments can be A system of forces and couple moments can be represented by a single force acting at a given point and represented by a single force acting at a given point and a single couple momenta single couple moment

To do so it is necessary that the force and couple To do so it is necessary that the force and couple moment system produce the same "external" effects of moment system produce the same "external" effects of translation and rotation of the body as their resultantstranslation and rotation of the body as their resultants

Two systems of forces and couple moments (system 1 Two systems of forces and couple moments (system 1 and system 2) are equivalent if:and system 2) are equivalent if:the sums of the forces are equal, the sums of the forces are equal,

((∑ ∑ FF))11 = (∑ = (∑ FF))22

and the sums of the moments about a point P are equal,and the sums of the moments about a point P are equal,((∑ ∑ MMpp))11 = (∑ = (∑ MMpp))22

EXAMPLES (pg 167 – 169)EXAMPLES (pg 167 – 169)

Special cases that occurSpecial cases that occur

Concurrent forces represented by a forceConcurrent forces represented by a force Coplanar forces represented by a forceCoplanar forces represented by a force Parallel forces represented by a forceParallel forces represented by a force

Representing a system by a wrenchRepresenting a system by a wrench

The wrench consists of the force The wrench consists of the force FFRR acting at a point P and acting at a point P and the parallel component the parallel component MMpp. The point P must be chosen so . The point P must be chosen so that the moment of that the moment of FFRR about P equals the normal about P equals the normal component component MMn,n, rrOPOP x x FFRR = = MMnn (SHOW)(SHOW)

EXAMPLES (pg 179 – 182)EXAMPLES (pg 179 – 182)

Simple Distributed LoadingSimple Distributed Loading

Plate supports the loading over its surfacePlate supports the loading over its surface Loading function, p = p(x) (Pa), is only a function of Loading function, p = p(x) (Pa), is only a function of

x since the pressure is uniform along the y axisx since the pressure is uniform along the y axis Multiply p = p(x) by the width a (m) of the plate, w Multiply p = p(x) by the width a (m) of the plate, w

= p(x) (N/m= p(x) (N/m22) a (m) = w(x) (N/m)) a (m) = w(x) (N/m) dF is acting on an infinitesimally thin element of dF is acting on an infinitesimally thin element of

width dx, dF = w(x) dx (= dA, in terms of the “area” width dx, dF = w(x) dx (= dA, in terms of the “area” A under the curve)A under the curve)

Can replace this coplanar parallel force system with Can replace this coplanar parallel force system with a single equivalent resultant forcea single equivalent resultant force

The magnitude of the resultant force is equal to the The magnitude of the resultant force is equal to the total area A under the loading diagram w=w(x)total area A under the loading diagram w=w(x)

L A

R AdAdxxwF )(

Location of the Resultant Force for Location of the Resultant Force for a Simple Distributed Loadinga Simple Distributed Loading

Location of to the line of action of Location of to the line of action of FFRR can be can be determined by equating the moments of the force determined by equating the moments of the force resultant and the force distribution about point Oresultant and the force distribution about point O

The resultant force has a line of action which The resultant force has a line of action which passes through the centroid C (the geometric passes through the centroid C (the geometric center) of the area defined by the distributed-center) of the area defined by the distributed-loading diagram w(x)loading diagram w(x)

Often the distributed-loading diagram is in the Often the distributed-loading diagram is in the shape of a rectangle, triangle, or some other shape of a rectangle, triangle, or some other simple geometric formsimple geometric form

EXAMPLES (pg 189 - 192)EXAMPLES (pg 189 - 192)

A

A

L

L

dA

xdA

dxxw

dxxxw

x)(

)(

x