Chapter 2 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.

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

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

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

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 )

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,

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

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

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

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