ch4 part 1

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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CHAPTER 4

FORCE SYSTEM

RESULTANTS

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Introduction

• Treatment of a body as a single particle is not always possible. In

general, the size of the body and the specific points of application of the

forces must be considered.

• Most bodies in elementary mechanics are assumed to be rigid, i.e., the

actual deformations are small and do not affect the conditions of

equilibrium or motion of the body.

• Current chapter describes the effect of forces exerted on a rigid body and

how to replace a given system of forces with a simpler equivalent system.

• moment of a force about a point

• moment of a force about an axis

• moment due to a couple

• Any system of forces acting on a rigid body can be replaced by an

equivalent system consisting of one force acting at a given point and one

couple.



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External and Internal Forces

• Forces acting on rigid bodies are

divided into two groups:

- External forces

- Internal forces

• External forces are shown in a

free-body diagram.

• If unopposed, each external force

can impart a motion of

translation or rotation, or both.



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Principle of Transmissibility: Equivalent Forces

• Principle of Transmissibility -

Conditions of equilibrium or motion are

not affected by transmitting a force

along its line of action.

NOTE: F and F’ are equivalent forces.

• Moving the point of application of

the force F to the rear bumper

does not affect the motion or the

other forces acting on the truck.

• Principle of transmissibility may

not always apply in determining

internal forces and deformations.



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Vector Product of Two Vectors

• Concept of the moment of a force about a point is

more easily understood through applications of

the vector product or cross product.

• Vector product of two vectors P and Q is defined

as the vector V which satisfies the following

conditions:

1. Line of action of V is perpendicular to plane

containing P and Q.

2. Magnitude of V is

3. Direction of V is obtained from the right-hand

rule.

sinQPV

• Vector products:

- are not commutative,

- are distributive,

- are not associative,

QPPQ

2121 QPQPQQP

SQPSQP



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Vector Products: Rectangular Components

• Vector products of Cartesian unit vectors,

0

0

0

kkikjjki

ijkjjkji

jikkijii

• Vector products in terms of rectangular

coordinates

kQjQiQkPjPiPV zyxzyx

kQPQP

jQPQPiQPQP

xyyx

zxxzyzzy

zyx

zyx

QQQ

PPP

kji

i j k i j k+

-

Memory Aid:

Memory Aid:



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Example: Vector Products (cross products)

Example:

a) Term by term

b) Using a determinant

c) Using a calculator (refer to the handout: "Cross and Dot Products“)

B A calculate then ,k5 - j i2 B and j4 - i3 A If



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Moment of a Force About a Point

• A force vector is defined by its magnitude and

direction. Its effect on the rigid body also depends

on it point of application.

• The moment of F about O is defined as

FrMO

• The moment vector MO is perpendicular to the

plane containing O and the force F.

• Any force F’ that has the same magnitude and

direction as F, is equivalent if it also has the same line

of action and therefore, produces the same moment.

• Magnitude of MO measures the tendency of the force

to cause rotation of the body about an axis along MO.

The sense of the moment may be determined by the

right-hand rule.

FdrFMO sin



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Moment of a Force About a Point (Scalar formulation)

• Two-dimensional structures have length and breadth but

negligible depth and are subjected to forces contained in

the plane of the structure.

• The plane of the structure contains the point O and the

force F. MO, the moment of the force about O is

perpendicular to the plane.

• If the force tends to rotate the structure clockwise, the

sense of the moment vector is into the plane of the

structure and the magnitude of the moment is negative.

• If the force tends to rotate the structure counterclockwise,

the sense of the moment vector is out of the plane of the

structure and the magnitude of the moment is positive.



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Varignon’s Theorem (Principle of Moments)

• The moment about a give point O of the

resultant of several concurrent forces is equal

to the sum of the moments of the various

moments about the same point O.

• Varignon’s Theorem makes it possible to

replace the direct determination of the

moment of a force F by the moments of two

or more component forces of F.

2121 FrFrFFr



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Rectangular Components of the Moment of a Force

(Vector formulation)

kyFxFjxFzFizFyF

FFF

zyx

kji

kMjMiMM

xyzxyz

zyx

zyxO

The moment of F about O,

kFjFiFF

kzjyixrFrM

zyx

O

,



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The moment of F about B,

FrM BAB

/

kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

/

zyx

BABABAB

FFF

zzyyxx

kji

M



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For two-dimensional structures,

xy

ZO

xyO

yFxF

MM

kyFxFM

xBAyBA

ZB

xBAyBAB

FyyFxx

MM

kFyyFxxM



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Demonstration – Calculating moments in 2D

Demonstration: Rope and wrench.

Illustrate how the angle and location of

rope affects the moment produced on the

bolt.

Example: Determine the moment about point A due to the 10

lb force applied to the wrench in each case below.

A

8"

10 lb

A

8"

10 lb

A

8"

10 lb

A

8"

10 lb

53.13 o

36.87 o

1)

2)

3)

4)



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Calculating moments in 2D – 4 methods

Calculation of moments in 2D – several approaches may be taken to calculate the

moment about point A due to the force shown below:

1. Using F, the perpendicular () component of the force. MA = (d)(F)

F d

A

F d

A

F

2. Using d, the perpendicular () component of the distance. MA = (d)(F)

F d

A

d



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Calculating moments in 2D – 4 methods

3. Using rectangular components for F and d. MA = (dx)(Fy) + (dy)(Fx)

F d

A

Fy

Fx

dy

dx

4. Using a cross product. This method is primarily used in 3D, but could be used in

2D problems also.

• determine the sign of each part by inspection

• CCW moments are +

• CW moments are -



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Example – Calculating moments in 2D

Example: A 100 lb force is applied to an 18 inch bar as shown below.

Determine the moment about point A due to the 100 lb force by:

1. Finding the perpendicular component of the force.

2. Finding the perpendicular component of the distance.

3. Using rectangular components of the force and the distance.

4. Using a cross product.

30

65

100 lb

A



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Example

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Example – Calculating moments in 2D

Example: (Problem 4-9 in Statics, 9th Ed. by Hibbeler)

Determine the moment due to the three forces:

a) about point A on the beam

b) about point B on the beam



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Example (Problem 4.1) – Also see Sample Problems 4.1-4.4

A 13.2 N force P is applied to the lever which

controls the auger of a snow blower. Determine the

moment of P about A when a is equal to 30°.



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Example

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Example

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Calculation of moments in 3D

The cross product is generally used for calculating moments in 3D problems.

A

where r a position vector from point A to any point along the line of action of F

Also note that M will have a direction that is perpendicular to the plane containing

r and F. The direction of

AM can be determined using the "right-hand rule."



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Example

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Example

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Example (Problem 4.22)

Before a telephone cable is strung, rope BAC is tied to the stake at B and is

passed over a pulley at A. Knowing that portion AC of the rope lies in a

plane parallel to the xy plane and that the tension T in the rope is 124 N,

determine the moment about O of the resultant force exerted on the pulley by

the rope. (Discuss Varignon’s Theorem).



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Example (Problem 4.24)

A wooden board AB, which is used as a temporary prop to support a small

roof, exerts at point A of the roof a 228 N force directed along BA.

Determine the moment about C of that force.

ch4 part 1

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