Week2 (1)2.2 vector operations Vector addition: (vector sum) Geometrical Method

Parallelogram Law Triangle construction In a “head-to-tail” fashion.

The method can be extended to n-D situation.

Form the parallelogram composed ofA and B. The diagnal is the vector sum. Vector addition is Commutative. Analytical method in Cartesian coordinate system.

= （Ax+ Bx ）i+（Ay +By ）j+ (Az+ Bz)k

Dot Product(sec.2.9)

Definition: A · B = AB cosq (A dot B)=A · B = Ax Bx + Ay By + Az Bz (in Cartesian coordinate system) =（ ）

A

B

A cosq

A ·u B (geometric meaning) : projection of A on B

The geometric meaning of A ·u B is the projection of A on the direction of B. If A · B = o , then A ┴B

Laws of operation1. Commutative law: A · B = B · A2. Multiplication by a scalar: a(A · B) = (aA) · B = A · (aB)3. Distributive law: A · (B + C) = (A · B) + (A · C)

Dot product can be used to determine 1. The angle between two lines 2. The component along specified direction

A

B

R

v

u

A B

Rv

1

u

Cross Product Sec.4.2 (be used to define the moment of a force about a point)

Definition: C = A ´ B = ( AB sinq ) uc (A cross B) Magnitude direction( right-handed role)The magnitude of A cross B is the area of the parallelogram containing A and B.

Proof : A ´ B=（Axi+Ay j+Azk）´（Bxi+By j+ Bzk）

Laws of operationCross product can be to define and calculate the moment of a force about a point.

Vector Calculus

Example 1:Determine the angle q between the tails of the two vectors. (Prob. 2-110) Z

Example 2:Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod. (Prob. 2-117)

x

y

4 m

3 m

6 m

2 m

3 m

r1

r2q

2

1. A ´ B = -B ´ A (Note: r ´ F F ´ r )2 c(A ´ B) = (cA) ´ B = A ´ (cB) 3 A ´ (B + D) = A ´ B + A ´ D

• unit vector along specified direction• vector’s component along specified direction

HOMEWOK 2.78, 2.118

Chapter 3 Equilibrium of a Particle

• Equilibrium Condition• Free-Body Diagram• 2-D Force System• 3-D Force System• Sec. 3.1 ~ 3.4

3.1 Equations of Equilibrium of a Particle

A

B

C

D

O

4 m

3 m

4 m

6 m

x

y

z

4 m

3

• Equilibrium or “static equilibrium”: acceleration = 0• å Fi = 0 All forces intersect at the particle and the forces are concurrent.• å moment (torque) = 0 (?)

3.2 Free-Body DiagramProcedures:

1. Establish the x,y,z axes in any suitable orientation.2. Label all the known & unknown (active & reactive) force magnitudes and

directions on the diagram.3. The known forces should be labeled with their magnitudes and directions.

The sense of a force having an unknown magnitude can be assumed. 3.3 3.4 Equilibrium Conditions of 2-D and 3-D Force Systems2-D: åFx=0 , åFy=0 Coplanar Force Systerm

3-D: åFx=0 , åFy=0 , åFz=0

Example 1:

Determine the required length of cord AC in Fig. 3-8a so that the 8-kg lamp is suspended in the position shown. The undeformed length of spring AB is l’AB = 0.4 m, and the spring has a stiffness of kAB = 300 N/m. (Exp. 3.4)

åFy= 0 Þ T sin(30o) = 78.4 Þ T = 156.8 (N)åFx= 0 Þ T cos(30o) = Fs Þ Fs = 135.8 (N)

Stretch = Fs/kAB = 135.8/300 = 0.453 (m)Total length = 0.453 + 0.4 = 0.853 (m)

lAC cos(30o) + 0.853 = 2lAC = 1.32 (m)Example 2:

Two electrically charges pith balls, each having a mass of 0.2 g, are suspended from light threads of equal length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r = 200 mm. (Prob. 3-11)

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Example 3:

The three cables are used to support the 800-N lamp. Determine the force developed in each cable for equilibrium. (Prob. 3-45)

Example 4:

Determine the stretch in each of the two springs required to hold the 20-kg crate in the equilibrium position shown. Each spring has an unstretched length of 2 m and a stiffness of k = 300 N/m. (Prob. 3-47)

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HOMEWOK 3.35 , 3.39, 3.53, 3.74

Chpater4: Force System Resultants

• Moment of a force• Moment of a couple• Reduction of a force and couple system• Reduction of a simple distributed loading

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Sec.4.1, Sec.4.3 - Sec.4.6

Sec.4.7 - Sec.4.10

Week 2。（2）Sec.4.1, Sec.4.3 - Sec.4.6

Definition: (Scalar Formulation & Vector Formulation)Moment of a force (“torque” or “moment”) – a measure of the tendency of the force to cause a body to rotate about the point or axis.4.1 Moment of a Force -Scalar FormulationShow the rotation of a door about the hinged z-axis.(Fx, Fy, Fz)

Coplanar force systems -- in 2-D situation: (conveniently viewed )

Magnitude: MO = F dWhere d is referred to as the moment arm or perpendicular distance from point O to the line of

action of the force.

Point O is a definite, about which moment are taken. So the name of the point can’t be

omitted.

Direction:Counterclockwise represents the positive sense. And clockwise, the negative sense.

In 2-D situation: The magnitude of the moment: the area of the parallelogram containing F and r.

The direction of the moment : Using the right- hand rule, the direction and the sense of the

moment vector are specified by the thumb, which points out of the page since the fingers follow

the curl. The moment always acts about an axis which is perpendicular to the plane containing F

and r.

So the vector form of the Moment of a Force can be written as Mo = r ´ F

Resultant Moment of a System of Coplanar Forces

If a system of forces lies in an x-y plane ,then the moment produced by each force about point O

will be directed along z axis. Consequently, the resultant moment MRO of the system can be

determined by simply adding the moment of all forces algebraically.

4.3 Moment of a Force- Vector FormulationIn 3-D situation:

Mo = r ´ F

Mo : moment of a force F about point Or : a position vector drawn from O to any point lying on the line of action of F

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Cartesine Vector Formulation.

Mo = r ´ F=

In established Cartesine Coordinate System, the position vector r and the force F can be expressed

as Cartesine Vector Forms.

Where r= rxi+ ryj +rzk F=Fx i＋Fy j+Fz k

And Mo= (ryFz- rz Fy)i+ (rzFx- rxFz)j +(rx Fy- ryFx) kThe physical meaning of these three components:

For example the k component ( Mo )z = (rx Fy- ryFx). Fz does not create a moment or tendency to cause turning about the z axis since this force is parallel to the z axis z axis.

rx Fy refers to the moment caused by Fy about the z axis.(positive)

ryFx refers to the moment caused by Fx about the z axis.(negative).

So (rx Fy- ryFx), the k component can be used to expressed the moment of the force F about the z axis.

Similar explanations can be obtained for i component and j component.

Principle of TransmissibilityF is a sliding vector and can act at any point along its line of action and still create the same moment about point O. (Condition?)

Mo = rA ´ F = rB ´ F

(How to proof it?)

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Resultant Moment of a System of Forces (to 4.7,4.9,4.8)

4.4 Principle of moment(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, i.e. Mo = r ´ F = r ´ (F1 + F2) = r ´ F1 + r ´ F2

Example 1:A 200-N force acts on the bracket shown in Fig. 4-19a. Determine the moment of the force about point A. (Example 4.6)

Solution I.

M = 200 ´ sin450 ´ 0.1 = 14.1 N·m

Solution II.

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Example 2:

The curved rod lies in the x-y plane and has a radius of 3 m. If a force of F = 80 N acts at its end

as shown, determine the moment of this force about point B. (Prob. 4-39)

4.5 Moment of a Force about a Specified Axis (The direction of the moment obviously along the Specified Axis) When a horizontal force F is applied to the handle of the flex-headed wrench, it creates a maximum moment, Fd. If the handle is not in the horizontal position, then the moment about the z axis is equal to the component of the moment along the z axis, Fd’ cosq.Ma：The effect of F in tending to rotate the body about the aa’ axis. The tendency is for rotation is measured by the moment component Ma .

The relation of Moment of a Force about a point and Moment of a Force about a Specified Axis

The component or projection of Mo onto the aa’ axis is the moment of a force about the specified axis aa’ on which the point O lies.Top view of Fig.4-14 (a) ry

(O’) O y rx Fy

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Fx

xMove the origin point of the coordinate system O to point O’, the top view remains unchanged. So the two k components (rx Fy- ryFx) k of the moments in both cases are the same. And hold for all points O’’ O’’’ ……… Considering that Fz can’t create the

tendency of the rotation about Z axis, the k component is the measure of the tendency of the force to cause a body to rotate about Z axis, represented by Mz(F), called the moment of F about Z axis. Conclusion : Mz(F)= (MO ) z

To determine the Ma , we first compute the moment of F about any arbitrary point O that lies on

the aa’ axis. And so the component or projection of Ma onto the aa’ axis is then Ma .

Ma=Mo cosθ= Ma ·ua

ua : unit vector that defines the direction encountered

Ma = Ma ua = [ua ×(r ´ F)] ua

Ma = å [ua ×(ri ´ Fi)] = ua × å (ri ´ Fi)

Example 3:

The chain AB exerts a force of 20 lb on the door at B. determine the magnitude of the moment of this force along the hinged axis x of the door. (Prob. 4-55)

Example 4:

If a torque or moment of 80 lb×in. is required to looser the bolt at A, determine the

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force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. (Prob. 4-65)

4.6 Moment of a Couple2 parallel forces:

• having same magnitude and opposite directions,

• separated by a perpendicular distance d.

• M = F d

A couple moment is a free vector. That is, a couple moment is independent of the location of reference point O. It can act at any point.

Equivalent Couples

Resultant Couple Moment.

Example 5:

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Replace the two couples acting on the pipe column in Fig. 4-32a by a resultant couple moment. (Example 4.13)

M1 = 60 I(N·m) MR = M1 + M2= 60 i + 22.5 j + 30 k (N·m)

M2 = 22.5 j +30 k

Homework4-39, 4-30, 4-41, 4-67,

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