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STATICS OF PARTICLES

Forces are vector quantities; they add according to theparallelogram law. The magnitude and direction of theresultant R of two forces P and Q can be determined eithergraphically or by trigonometry.

P

R

QA

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A

Q

P

F

Any given force acting on a particle can be resolved into two or more components, i.e.., it can be replaced by two or more forces which have the same effect on the particle.

A force F can be resolvedinto two components Pand Q by drawing aparallelogram which hasF for its diagonal; thecomponents P and Qare then represented bythe two adjacent sidesof the parallelogramand can be determinedeither graphically or bytrigonometry.

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x

y

Fx = Fx i

Fy = Fy j

F

i

j

A force F is said to have been resolved into two rectangularcomponents if its components are directed along the coordinateaxes. Introducing the unit vectors i and j along the x and y axes,

F = Fx i + Fy j

Fx = F cos Fy = F sin

tan = Fy

Fx

F = Fx + Fy2 2

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When three or more coplanar forces act on a particle, the rectangular components of their resultant R can be obtained by adding algebraically the corresponding components of the given forces.

Rx = Rx Ry = Ry

The magnitude and direction of R can be determined from

tan = Ry

RxR = Rx + Ry

2 2

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x

y

z

A

B

C

D

E

OFx

Fy

Fz

F

x

y

z

A

B

C

D

E

OFx

Fy

Fz

F

x

x

y

z

A

B

C

D

E

OFx

Fy

Fz

F

y

z

A force F in three-dimensional space can be resolved into components

Fx = F cos x Fy = F cos y

Fz = F cos z

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x

y

z

Fy j

Fx i

Fz k

(Magnitude = 1)

cos x i

cos z k

cos y j F = F

The cosines ofx , y , and z

are known as thedirection cosines of the force F. Using the unit vectors i , j, and k, we write

F = Fx i + Fy j + Fz k

or

F = F (cosx i + cosy j + cosz k )

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x

y

z

Fy j

Fx i

Fz k

(Magnitude = 1)

cos x i

cos z k

cos y j

F = F

= cosx i + cosy j + cosz k

Since the magnitude of is unity, we have

cos2x + cos2y

+ cos2z = 1

F = Fx + Fy + Fz2 2 2

cosx =Fx

Fcosy =

Fy

Fcosz =

Fz

F

In addition,

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x

y

z

M (x1, y1, z1)

N (x2, y2, z2)

dx = x2 - x1

dz = z2 - z1 < 0

dy = y2 - y1

A force vector Fin three-dimensionsis defined by itsmagnitude F andtwo points M and N along its line ofaction. The vectorMN joining points and N is

MN = dx i + dy j + dz k

= = ( dx i + dy j + dz k ) MNMN

1d

F

The unit vector along the line of action of the force is

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x

y

z

M (x1, y1, z1)

N (x2, y2, z2)

dx = x2 - x1

dz = z2 - z1 < 0

dy = y2 - y1

A force F is defined as theproduct of F and. Therefore,

F = F = ( dx i + dy j + dz k ) Fd

d = dx + dy + dz

22 2

From this it follows that

Fx =Fdx

dFy =

Fdy

dFz =

Fdz

d

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When two or more forces act on a particle in three-dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces.

The particle is in equilibrium when the resultant of all forces acting on it is zero.

Rx = Fx

Ry = Fy

Rz = Fz

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Fx = 0 Fy = 0 Fz = 0

In two-dimensions , only two of these equations are needed

To solve a problem involving a particle in equilibrium, draw a free-body diagram showing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium are

Fx = 0 Fy = 0

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q

V = P x Q

P

Q

The vector product of twovectors is defined as

V = P x QThe vector product of P and Q forms a vector which is

perpendicular to both P and Q, of magnitude

V = PQ sin This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through which brings vector P in line with vector Q. The three vectors P,Q, and V - taken in that order - form a right-hand triad. It follows that

Q x P = - (P x Q)

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ik

j It follows from the definition of the vectorproduct of two vectors that the vectorproducts of unit vectors i, j, and k are

i x i = j x j = k x k = 0

i x j = k , j x k = i , k x i = j , i x k = - j , j x i = - k , k x j = - i

The rectangular components of the vector product V of twovectors P and Q are determined as follows: Given

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k

The determinant containing each component of P and Q isexpanded to define the vector V, as well as its scalar components

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P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k

V = P x Q =i

Px

Qx

jPy

Qy

kPz

Qz

= Vx i + Vy j + Vz k

where

Vx = Py Qz - Pz Qy

Vy = Pz Qx - Px Qz

Vz = Px Qy - Py Qx

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Od A

F

Mo

r

The moment of force F about point O is defined as the vector product

MO = r x F

where r is the positionvector drawn from point O to the point of application of the force F. The angle between the lines of actionof r and F is .

The magnitude of the moment of F about O can be expressed as

MO = rF sin = Fd

where d is the perpendicular distance from O to the line of actionof F.

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x

y

z

Fx i

Fz k

Fy j

x i

y j

z k

O

A (x , y, z )

r

The rectangular components of themoment Mo of a force F are determined byexpanding thedeterminant of r x F.

Mo = r x F =ixFx

jyFy

kzFz

= Mx i + My j + Mzk

where Mx = y Fz - z Fy My = zFx - x Fz

Mz = x Fy - y Fx

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x

y

Fx i

Fz k

Fy j

O

r

In the more general case of the moment about an arbitrary point B of a force F applied at A, we have

MB = rA/B x F =i

xA/B

Fx

jyA/B

Fy

kzA/B

Fz

where

z

B (x B, yB, z B)

A (x A, yA, z A)

rA/B = xA/B i + yA/B j + zA/B k

and xA/B = xA- xB yA/B = yA- yB zA/B = zA- zB

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x

y

Fx i

FFy j

O

MB = (xA- xB )Fy + (yA- yB ) Fx

z

B

A

(xA - xB ) i

rA/B(yA - yB ) j

In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalar

MB = MB k

The right-hand rule is useful for defining the direction of themoment as either into or out of the plane (positive or negativek direction).

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P

QThe scalar product of two vectors P and Q is denoted as P Q ,and is defined as

The scalar product of P and Q is expressed in terms of the rectangular components of the two vectors as

P Q = PQ cos

P Q = Px Qx + Py Qy + Pz Qz

where is the angle between the two vectors

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x

y

z

O

L

A

xP

z

y

The projection of a vectorP on an axis OL can be obtained by forming the scalar product of P and the unit vectoralong OL.

POL = P

Using rectangular components,

POL = Px cos x + Py cos y + Pz cos z

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The mixed triple product of three vectors S, P, and Q is

S (P x Q ) =Sx

Px

Qx

Sy

Py

Qy

Sz

Pz

Qz

The elements of the determinant are the rectangular components of the three vectors.

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x

F

O

The moment of a force F about an axis OL is the projection OC on OL of the moment MO of the force F. This can be written as a mixed triple product.

z

A (x, y, z)

yMO

L

C

r

x

xFx

y

yFy

z

zFz

MOL =MO =(r x F) =

x, y , z = direction cosines of axis OLx, y , z = components of rFx, Fy , Fz = components of F

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dF

- FM

Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.

The moment of a couple is independent of the point about which it is computed; it is a vector M perpendicular to the plane of the couple and equal in magnitude to the product Fd.

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Two couples having the same moment M are equivalent (they have the same effect on a given rigid body).

x

y

z

dF

- F

x

y

zO O

M

(M = Fd)

x

y

z

O Mx

My

Mz

M

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Any force F acting at a point A of a rigid body can be replaced by a force-couple system at an arbitrary point O, consisting of the force F applied at O and a couple of moment MO equal to the moment about point O of the force F in its original position. The force vector F and the couple vector MO are alwaysperpendicular to each other.

O

Ar

F

O

A

FMO

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O

A1r1

F1

O

M1

r2 A2

F2

r3

A3

F3

F1

M2M3

F2

F3 R

MRO

O

Any system of forces can be reduced to a force-couple system at a given point O. First, each of the forces of the system is replaced by an equivalent force-couple system at O. Then all of the forces are added to obtain a resultant force R, and all of couples are added to obtain a resultant couple vectorMO. In general, the resultant force R and the couple vector MO

will not be perpendicular to each other.

R R

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O

A1r1

F1

r2 A2

F2

r3

A3

F3

R

MRO

O

As far as rigid bodies are concerned, two systems of forces, F1, F2, F3 . . . , and F’1, F’2, F’3 . . . , are equivalent if, and only if,

F = F’ and

Mo = Mo’

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O

A1r1

F1

r2 A2

F2

A3

F3

R

M RO

O

This is the case for systems consisting of either (a) concurrent forces, (b) coplanar forces, or (c) parallel forces.

If the resultant force and couple are directed along the same line, the force-couple system is termed a wrench.

R

If the resultant force R and the resultant couple vector MO areperpendicular to each other, the force-couple system at O can be further reduced to a single resultant force.