Transcript Chapter 4 - Equilibrium of Particle

Chapter 3:
Equilibrium of a Particle
Engineering Mechanics: Statics
Chapter Objectives
To introduce the concept of the free-body
diagram for a particle.
 To show how to solve particle equilibrium
problems using the equations of equilibrium.

Chapter Outline
 Condition
for the Equilibrium of a
Particle
 The Free-Body Diagram
 Coplanar Systems
 Three-Dimensional Force Systems
3.1 Condition for the
Equilibrium of a Particle
Particle at equilibrium if
- At rest
- Moving at constant a constant velocity
 Newton’s first law of motion
∑F = 0
where ∑F is the vector sum of all the
forces acting on the particle

3.1 Condition for the
Equilibrium of a Particle
Newton’s second law of motion
∑F = ma
 When the force fulfill Newton's first law
of motion,
ma = 0
a=0
therefore, the particle is moving in
constant velocity or at rest

3.2 The Free-Body Diagram
Best representation of all the unknown
forces (∑F) which acts on a body
 A sketch showing the particle “free” from
the surroundings with all the forces acting
on it
 Consider two common connections in this
subject – Spring
– Cables and Pulleys

3.2 The Free-Body Diagram

Spring
- Linear elastic spring: change in length is
directly proportional to the force acting on it
- spring constant or stiffness k:
defines the elasticity of
the spring
- Magnitude of force when spring
is elongated or compressed
F = ks
3.2 The Free-Body Diagram

Spring
where s is determined from the difference in
spring’s deformed length l and its
undeformed length lo
s = l - lo
- If s is positive, F “pull”
onto the spring
- If s is negative, F “push”
onto the spring
3.2 The Free-Body Diagram
Example
Given lo = 0.4m and k = 500N/m
To stretch it until l = 0.6m, A force, F = ks
=(500N/m)(0.6m – 0.4m) = 100N is needed
To compress it until l = 0.2m,
A force, F = ks
=(500N/m)(0.2m – 0.4m)
= -100N is needed
3.2 The Free-Body Diagram

Cables and Pulley
- Cables (or cords) are assumed to have
negligible weight and they cannot stretch
- A cable only support tension or pulling force
- Tension always acts in the
direction of the cable
- Tension force in a continuous
cable must have a constant
magnitude for equilibrium
3.2 The Free-Body Diagram
 Cables
and Pulley
- For any angle θ, the cable is
subjected to
a constant tension T
throughout its length
3.2 The Free-Body Diagram
Procedure for Drawing a FBD
1. Draw outlined shape
- Isolate particle from its surroundings
2. Show all the forces
- Indicate all the forces
- Active forces: set the particle in motion
- Reactive forces: result of constraints and
supports that tend to prevent motion
3.2 The Free-Body Diagram
Procedure for Drawing a FBD
3. Identify each forces
- Known forces should be labeled with
proper magnitude and direction
- Letters are used to represent
magnitude and directions of unknown
forces
3.2 The Free-Body Diagram




A spool is having a weight
W which is suspended from
the crane bottom
Consider FBD at A since
these forces act on the ring
Cables AD exert a resultant
force of W on the ring
Condition of equilibrium is
used to obtained TB and TC
3.2 The Free-Body Diagram





The bucket is held in
equilibrium by the cable
Force in the cable =
weight of the bucket
Isolate the bucket for
FBD
Two forces acting on
the bucket, weight W
and force T of the cable
Resultant of forces = 0
W=T
3.2 The Free-Body Diagram
Example 3.1
The sphere has a mass of 6kg and is
supported. Draw a free-body diagram of
the
sphere, the cord
CE and the knot at C.
3.2 The Free-Body Diagram
Solution
FBD at Sphere
 Two forces acting,
weight and the
force on cord CE.
 Weight of 6kg
(9.81m/s2) = 58.9N
3.2 The Free-Body Diagram
Solution
Cord CE
 Two forces acting, force
of the sphere and force
of the knot
 Newton’s Third Law: FCE
is equal but opposite
 FCE and FEC pull the cord
in tension
 For equilibrium, FCE =
FEC
3.2 The Free-Body Diagram
Solution
FBD at Knot
 Three forces acting, force by cord CBA, cord CE
and spring CD
 Important to know that
the weight of the sphere
does not act directly on
the knot but subjected to
by the cord CE
3.3 Coplanar Systems



A particle is subjected to coplanar forces in
the x-y plane
Resolve into i and j components for
equilibrium
∑Fx = 0
∑Fy = 0
Scalar equations of equilibrium
require that the algebraic sum
of the x and y components to
equal o zero
3.3 Coplanar Systems

Scalar Notation
- Sense of direction = an algebraic sign
that corresponds to the arrowhead
direction of the component along each axis
- For unknown magnitude, assume
arrowhead sense of the force
- Since magnitude of the force is always
positive, if the scalar is negative, the force
is acting in the opposite direction
3.3 Coplanar Systems
Example
Consider the free-body diagram of the
particle subjected to two forces


Assume unknown force F acts to the right for
equilibrium
∑Fx = 0 ;
+ F + 10N = 0
F = -10N
Force F acts towards the left for equilibrium
3.3 Coplanar Systems


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The chain exerts three
forces on the ring at A.
The ring will not move, or
will move with constant
velocity, provided the
summation of the forces
along the y axis is zero
With any force known, the
magnitude of other two
forces are found by
equations of equilibrium
3.3 Coplanar Systems
 Procedure
for Analysis
1. Free-Body Diagram
- Establish the x, y axes in any suitable
orientation
- Label all the unknown and known
forces magnitudes and directions
- Sense of the unknown force can be
assumed
3.3 Coplanar Systems
 Procedure
for Analysis
2) Equations of Equilibrium
- Given two unknown with a spring, apply
F = ks
to find spring force using deformation of
spring
- If the solution yields a negative result,
the sense of force is the reserve of that
shown in the free-body diagram
3.3 Coplanar Systems
 Procedure
for Analysis
2) Equations of Equilibrium
- Apply the equations of equilibrium
∑Fx = 0
∑Fy = 0
- Components are positive if they are
directed along the positive negative axis
and negative, if directed along the
negative axis
3.3 Coplanar Systems
Example 3.2
Determine the tension in
cables AB and AD for
equilibrium of the 250kg
engine.
3.3 Coplanar Systems
Solution
FBD at Point A
- Initially, two forces acting, forces
of cables AB and AD
- Engine Weight
= (250kg)(9.81m/s2)
= 2.452kN supported by cable CA
- Finally, three forces acting, forces
TB and TD and engine weight
on cable CA
3.3 Coplanar Systems
Solution
+→ ∑Fx = 0;
+↑ ∑Fy = 0;
Solving,
TBcos30° - TD = 0
TBsin30° - 2.452kN = 0
TB = 4.90kN
TD = 4.25kN
*Note: Neglect the weights of the cables since they
are small compared to the weight of the engine
3.3 Coplanar Systems
Example 3.3
If the sack at A has a weight
of 20N (≈ 2kg), determine
the weight of the sack at B
and the force in each cord
needed to hold the system in
the equilibrium position
shown.
3.3 Coplanar Systems
Solution
FBD at Point E
- Three forces acting,
forces of cables EG
and EC and the
weight of the sack on
cable EA
3.3 Coplanar Systems
Solution
+→ ∑Fx = 0;
+↑ ∑Fy = 0;
Solving,
TEGsin30° - TECcos45° = 0
TEGcos30° - TECsin45° - 20N = 0
TEC = 38.6kN
TEG = 54.6kN
*Note: use equilibrium at the ring to determine
tension in CD and weight of B with TEC known
3.3 Coplanar Systems
Solution
FBD at Point C
- Three forces acting, forces by cable
CD
and EC (known) and
weight of sack B on
cable CB
3.3 Coplanar Systems
Solution
+→ ∑Fx = 0;
+↑ ∑Fy = 0;
Solving,
38.6cos30° - (4/5)TCD = 0
(3/5)TCD – 38.6sin45°N – WB = 0
TCD = 34.1kN
WB = 47.8kN
*Note: components of TCD are proportional to the slope of the cord by the 34-5 triangle
3.4 Three-Dimensional
Force Systems

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
For particle equilibrium
∑F = 0
Resolving into i, j, k components
∑Fxi + ∑Fyj + ∑Fzk = 0
Three scalar equations representing algebraic
sums of the x, y, z forces
∑Fxi = 0
∑Fyj = 0
∑Fzk = 0
3.4 Three-Dimensional
Force Systems

Make use of the three scalar equations
to solve for unknowns such as angles
or magnitudes of forces
3.4 Three-Dimensional Force
Systems


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Ring at A subjected to
force from hook and
forces from each of the
three chains
Hook force = weight of
the electromagnet and
the load, denoted as W
Three scalars equations
applied to FBD to
determine FB, FC and FD
3.4 Three-Dimensional
Force Systems
 Procedure
for Analysis
Free-body Diagram
- Establish the z, y, z axes in any suitable
orientation
- Label all known and unknown force
magnitudes and directions
- Sense of a force with unknown
magnitude can be assumed
3.4 Three-Dimensional
Force Systems

Procedure for Analysis
Equations of Equilibrium
- Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0 when forces
can be easily resolved into x, y, z components
- When geometry appears difficult, express each
force as a Cartesian vector. Substitute vectors
into ∑F = 0 and set i, j, k components = 0
- Negative results indicate that the sense of the
force is opposite to that shown in the FBD.
3.4 Three-Dimensional
Force Systems
Example 3.4
A 90N load is suspended from the hook. The
load is supported by two cables and a spring
having a stiffness k = 500N/m.
Determine the force in the
cables and the stretch of the
spring for equilibrium. Cable
AD lies in the x-y plane and
cable AC lies in the x-z plane.
3.4 Three-Dimensional
Force Systems
Solution
FBD at Point A
- Point A chosen as the forces are
concurrent at this point
3.4 Three-Dimensional
Force Systems
Solution
Equations
∑Fx = 0;
∑Fy = 0;
∑Fz = 0;
Solving,
of Equilibrium,
FDsin30° - (4/5)FC = 0
-FDcos30° + FB = 0
(3/5)FC – 90N = 0
FC = 150N
FD = 240N
FB = 208N
3.4 Three-Dimensional
Force Systems
Solution
For the stretch of the spring,
FB = ksAB
208N = 500N/m(sAB)
sAB = 0.416m
3.4 Three-Dimensional
Force Systems
Solution
Equations of Equilibrium
Expressing each forces in Cartesian
vectors,
FB = FBi
FC = FCcos120°i + FCcos135°j –
FCcos60°k
FD = -0.333FDi + 0.667FDj + 0.667FDk
W = -981k
3.4 Three-Dimensional
Force Systems
Solution
For equilibrium,
∑F = 0;
FB + FC + FD + W = 0
FBi + FCcos120°i + FCcos135°j – FCcos60°k
-0.333FDi + 0.667FDj + 0.667FDk - 981k
=0
∑Fx = 0; FB + FCcos120° - 0.333FD = 0
∑Fy = 0; FCcos135° + 0.667FD = 0
∑Fz = 0; FCcos60° + 0.667FD - 981 = 0
3.4 Three-Dimensional
Force Systems
Solution
Solving,
FB = 693.7N
FC = 813N
FD = 693.7N
For the stretch of the spring,
F B = ks
693.7N = 1500s
s = 0.462m