Engineering Mechanics: Statics
Introduction –Equilibrium of Rigid Bodies
• 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. As a result, the rigid bodies can translate as
well as rotate
• We first consider the effect of forces exerted on a rigid body and 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.
3-1
Engineering Mechanics: Statics
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.
3-2
Engineering Mechanics: Statics
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 V = P Q sin θ
3. Direction of V is obtained from the right-hand
rule.
• Vector products:
- are not commutative, Q × P = −( P × Q )
- are distributive,
P × (Q1 + Q2 ) = P × Q1 + P × Q2
- are not associative, ( P × Q ) × S ≠ P × (Q × S )
3-3
Engineering Mechanics: Statics
Vector Products: Rectangular Components
• Vector products of Cartesian unit vectors,
r r r r
r r
r r
i ×i = 0
j × i = −k k × i = j
r r
r r r
r r
r
i× j =k
j× j =0
k × j = −i
r
r
r r
r
r v
r
i ×k = − j j ×k = i
k ×k = 0
• Vector products in terms of rectangular
coordinates
r
r
r
r
r
r
r
V = (Px i + Py j + Pz k )× (Q x i + Q y j + Q z k )
r
r
= (Py Qz − Pz Q y )i + ( Pz Qx − Px Q z ) j
r
+ (Px Q y − Py Qx )k
r
r
r
i
j
k
= Px Py Pz
Qx Q y Qz
3-4
Engineering Mechanics: Statics
Moment of a Force About a Point
• The moment of F about O is defined as
MO = r × F
• The moment vector MO is perpendicular to the
plane containing O and the force F.
• Magnitude of MO measures the tendency of the force
to cause rotation of the body about an axis along MO.
M O = rF sin θ = Fd
The sense of the moment may be determined by the
right-hand rule.
3-5
Engineering Mechanics: Statics
Rectangular Components of the Moment of a Force
The moment of F about O,
r
r
r r
r r
r
M O = r × F , r = xi + yj + zk
r
r
r
r
F = Fx i + Fy j + Fz k
r
r
r
r
M O = M xi + M y j + M z k
r
i
r
j
r
k
= x
y
z
Fx
Fy
Fz
r
r
r
= yFz − zFy i + ( zFx − xFz ) j + xFy − yFx k
(
)
(
)
3-6
Engineering Mechanics: Statics
Scalar Product of Two Vectors
• The scalar product or dot product between
two vectors P and Q is defined as
r r
P • Q = PQcosθ (scalar result)
• Scalar products:
- are commutative,
- are distributive,
- are not associative,
r r r r
P•Q = Q• P
r r r
r r r r
P • (Q1 + Q2 ) = P • Q1 + P • Q2
r r r
(P • Q )• S = undefined
• Scalar products with Cartesian unit components,
r
r
r r
r
r
r
r
P • Q = (Px i + Py j + Pz k )• (Qx i + Q y j + Qz k )
r r
r r
r r
r r
i •i =1 j • j =1 k •k =1 i • j = 0
r r
v r
j •k = 0 k •i = 0
r r
P • Q = Px Qx + Py Q y + Pz Qz
r r
P • P = Px2 + Py2 + Pz2 = P 2
3-7
Engineering Mechanics: Statics
Scalar Product of Two Vectors: Applications
• Angle between two vectors:
r r
P • Q = PQ cosθ = Px Qx + Py Q y + Pz Qz
cosθ =
Px Qx + Py Q y + Pz Qz
PQ
• For an axis defined by a unit vector:
r r
POL = P • λ
= Px cosθ x + Py cosθ y + Pz cosθ z
3-8
Engineering Mechanics: Statics
Moment of a Force About a Given Axis
• Moment MO of a force F applied at the point A
about a point O,
r
r r
MO = r × F
• Scalar moment MOL about an axis OL is the
projection of the moment vector MO onto the
axis,
r r
r r r
M OL = λ • M O = λ • (r × F )
3-9
Engineering Mechanics: Statics
Moment of a Couple
• Two forces F and -F having the same magnitude,
parallel lines of action, and opposite sense are said
to form a couple.
• Moment of the couple,
r r r r
r
M = rA × F + rB × (− F )
r
r r
= (rA − rB ) × F
r r
= r ×F
M = rF sin θ = Fd
• The moment vector of the couple is
independent of the choice of the origin of the
coordinate axes, i.e., it is a free vector that can
be applied at any point with the same effect.
3 - 10
Engineering Mechanics: Statics
Couples Can Be Represented by Vectors
• A couple can be represented by a vector with magnitude
and direction equal to the moment of the couple.
• Couple vectors obey the law of addition of vectors.
• Couple vectors are free vectors, i.e., the point of application
is not significant.
• Couple vectors may be resolved into component vectors.
3 - 11
Engineering Mechanics: Statics
Resolution of a Force Into a Force at O and a Couple
• Force vector F can not be simply moved to O without modifying its
action on the body.
• Attaching equal and opposite force vectors at O produces no net
effect on the body.
• The three forces may be replaced by an equivalent force vector and
couple vector, i.e, a force-couple system.
3 - 12
Engineering Mechanics: Statics
System of Forces: Reduction to a Force and Couple
• A system of forces may be replaced by a collection of
force-couple systems acting a given point O
• The force and couple vectors may be combined into a
resultant force vector and a resultant couple vector,
r
r
rR
r r
R = ∑F
M O = ∑ (r × F )
• The force-couple system at O may be moved to O’
with the addition of the moment of R about O’ ,
rR
rR r r
M O' = M O + s × R
• Two systems of forces are equivalent if they can be
reduced to the same force-couple system.
3 - 13