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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
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
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