MEC30: STATICS OF RIGID BODIES
BY: ENGR. LAMBERTO B. MARCIAL, JR.
BASIC PRINCIPLES OF MECHANICS
MEC30: STATICS OF RIGID BODIES
BASIC PRINCIPLES OF MECHANICS
Mechanics
• is a branch of the physical sciences that is concerned with
the state of rest or motion of bodies that are subjected to the
action of forces.
Three Branches of Mechanics
• Rigid-Body Mechanics
• Deformable-Body Mechanics
• Fluid Mechanics
BASIC PRINCIPLES OF MECHANICS
Rigid-Body Mechanics
• Is divided into two areas:
– Statics
– Dynamics
Statics
Dynamics
• deals with the effects and
distribution of forces on rigid
bodies which are and remain at
rest.
• is concerned with the accelerated
motion of bodies.
BASIC PRINCIPLES OF MECHANICS
Rigid Body
• is a combination of a large number of particles in which all the
particles remain at a fixed distance from one another, both before
and after applying a load.
Particle
• is a representative of the body (whole) which has a mass but
the size can be neglected
BASIC PRINCIPLES OF MECHANICS
External or Applied Force
• is anything which changes or tends to change the state of
motion of a body.
Concurrent Forces
• are forces whose lines of action are passing through one
common point.
Parallel Forces
• are forces in the same direction whose lines of action never
meet.
BASIC PRINCIPLES OF MECHANICS
Concurrent Forces
Parallel Forces
BASIC PRINCIPLES OF MECHANICS
Resultant Force
• is the equivalent single force that represents a system of
forces.
Equilibrant Force
• is also known as the balancing force which has the same
magnitude as the resultant force but in opposite direction.
BASIC OPERATIONS WITH FORCE SYSTEM
Law of Inertia (Newton’s First Law of Motion)
• An object at rest stays at rest and an object in motion stays in
motion with the same speed and in the same direction unless
acted upon by an unbalanced force.
Action and Reaction Law (Newton’s Third Law of Motion)
• for every action, there is always an opposite and equal
reaction.
• action and reaction forces are equal but opposite in direction.
BASIC OPERATIONS WITH FORCE SYSTEM
Superposition Law or Law of Transmissibility
• It states that the state of rest or motion of a rigid body is unaltered
if a force acting on the body is replaced by another force of the
same magnitude and direction but acting anywhere on the body
in the line of action of the replaced force.
BASIC OPERATIONS WITH FORCE SYSTEM
Resultant of Force Systems
• the effect of a system of forces on a body is usually expressed in
terms of a resultant.
Note: A force is a vector quantity which means it is characterized by
its magnitude and direction.
VECTOR OPERATIONS
Vector Addition
• All vector quantities obey the parallelogram law of addition .
Parallelogram Law
• States that the resultant of two
forces is the diagonal of the
parallelogram formed on the two
vector of those forces.
PARALLELOGRAM LAW
Determining Resultant using Parallelogram Law
• The magnitude of the resultant force can be determined using
law of cosines, and its direction using the law of sines. The
magnitudes of two force components are determined from law of
sines.
PARALLELOGRAM LAW
Sample Problem
Determine the resultant and direction of the resultant.
Answer: 212.6 N, 54.8o above positive x-axis
PARALLELOGRAM LAW
Sample Problem
At what angle must the 800lb force be applied in order that the
resultant the resultant R of the two forces has a magnitude of
2000lb? For this condition, determine the angle β between R and
the vertical force.
Answer: β = 18.19o
COMPONENT OF A FORCE
At a given angle:
Fy = Fsinθ
F
θ
Fx = Fcosθ
COMPONENT OF A FORCE
At a given slope:
Fy = F(b/c)
F
c
c2 = a2 + b2
b
a
Fx = F(a/c)
COMPONENT OF A FORCE
Sign Convention:
Fy
F
c
a
Fx
b
positive (+)
negative (-)
COMPONENT OF A FORCE
Sample Problem
Determine the resultant and direction of the resultant.
Answer: 212.6 N, 54.8o above positive x-axis
COMPONENT OF A FORCE
Sample Problem
At what angle must the 800lb force be applied in order that the
resultant the resultant R of the two forces has a magnitude of
2000lb? For this condition, determine the angle β between R and
the vertical force.
Answer: β = 18.19o
DETERMINING THE RESULTANT FORCE
Sample Problem
Determine the Resultant along x and along y, the Magnitude of the
Resultant and Direction of the Resultant of the system shown.
Answer: R = 567.28 N,
θ = 38.1o above +x-axis
DETERMINING THE RESULTANT FORCE
Sample Problem
Determine the magnitude of the resultant force and its direction
measured from the positive x-axis.
Answer: R = 1040.82 N,
θ = 56.42o above +x-axis
DETERMINING THE RESULTANT FORCE
Determining θ
y
(-Rx, +Ry)
(+Rx, +Ry)
θ
θ
θ
θ
(-Rx, -Ry)
PROVIDED,
x
(+Rx, -Ry)
0o < θ < 90o
DETERMINING THE RESULTANT FORCE
Sample Problem
Determine the magnitude of the resultant force and its direction
measured from the positive x-axis.
Answer: R = 545.79 N,
θ = 252.64o from +x-axis
SYSTEMS IN EQUILIBRIUM
MEC30: STATICS OF RIGID BODIES
SYSTEMS IN EQUILIBRIUM
STATIC EQUILIBRIUM
• When forces acting on an object which is at rest are
balanced.
• Even if there are forces exerted toward the
object, these forces are balanced. Thus, the
forces will not incur any movement. Object
remains at rest and is at static equilibrium.
• At static equilibrium, the resultant is equal
to zero (Fr= 0).
SYSTEMS IN EQUILIBRIUM
CONDITIONS UNDER STATIC EQUILIBRIUM
• TRANSLATIONAL EQUILIBRIUM – the sum of forces acting
on a body is equal to zero.
•
•
•
•
Ftotal = F1 + F2 + F3 + …… + Fn = 0
Fx = F1x + F2x + F3x + …… + Fnx = 0
Fy = F1y + F2y + F3y + …… + Fny = 0
Fz = F1z + F2z + F3z + …… + Fnz = 0
SYSTEMS IN EQUILIBRIUM
CONDITIONS UNDER STATIC EQUILIBRIUM
• ROTATIONAL EQUILIBRIUM – the sum of the moment forces
or torque acting on a body is zero.
Mtotal = M1 + M2 + M3 + …… + Mn = 0
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
Determine the tension in cables BA and BC necessary to
support the 60kg cylinder.
FBC = 475.66 N
FBA = 420.43 N
We can also use the Force Triangle
to solve for FBC and FBA
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
If the mass of cylinder C is 40 kg, determine the mass of cylinder A
in order to hold the assembly in the position shown.
mA = 20kg
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
Determine the maximum weight of concrete block and the contact
force between the block and the incline that can be supported by force
P = 500N as shown in the figure. .Note: Assume that the concrete block
is resting on a smooth surface.
W = 642.79 N
CF = 939.69 N
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
Determine the reactions at A, B, C, and D.
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
Determine the reactions at A, B, C, and D.
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
Determine the reactions at A, B, C, and D.
SYSTEMS IN EQUILIBRIUM
SAMPLE PROBLEM
Determine the reactions at A, B, C, and D.
RA = 347.44 kN
RB = 600 kN
RC = 400.89 kN
RD = 347.44 kN
SEATWORK
SEATWORK
Resultant of a Force
Determine the resultant of the following concurrent forces
systems:
SEATWORK
SYSTEMS IN EQUILIBRIUM
Determine the force P and its angle of inclination, θ, that will
make the system in equilibrium.
SEATWORK
SYSTEMS IN EQUILIBRIUM
If Block D weighs 300lb and Block B weighs 275lb, determine the
weight of Block C and the angle θ for equilibrium.