Static Equilibrium
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Equilibrium implies the object is at rest
(static) or its center of mass moves with a
constant velocity (dynamic)
Static equilibrium is a common situation in
engineering
Principles involved are of particular interest to
civil engineers, architects, and mechanical
engineers
Conditions for Equilibrium
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The net force equals zero
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The net torque equals zero
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Torque
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τ=rxF
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Use the right-hand rule to
determine the direction of
the torque
The tendency of the force
to cause a rotation about
O depends on F and the
moment arm d
ΣF = 0
If the object is modeled as a particle, then this is
the only condition that must be satisfied
Στ = 0
This is needed if the object cannot be modeled as
a particle
Rotational Equilibrium
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Need the angular acceleration of the object to
be zero
For rotation, Στ = Iα
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For rotational equilibrium, Στ = 0
This must be true for any axis of rotation
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Equilibrium Summary
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There are two necessary conditions for equilibrium
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The resultant external force must equal zero:
ΣF = 0
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This is a statement of translational equilibrium
The acceleration of the center of mass of the object must be zero
when viewed from an inertial frame of reference
Equilibrium Equations
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This is a statement of rotational equilibrium
The angular acceleration must equal zero
Axis of Rotation for Torque Equation
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The net torque is about an axis through any
point in the xy plane
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The choice of an axis is arbitrary
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If an object is in translational equilibrium and
the net torque is zero about one axis, then
the net torque must be zero about any other
axis
These are called coplanar forces since they lie
in the same plane
There are three resulting equations
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The resultant external torque about any axis must
be zero: Στ = 0
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We will restrict the applications to
situations in which all the forces lie in the
xy plane
ΣFx = 0
ΣFy = 0
Στ z = 0
Center of Gravity…
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All the various
gravitational forces
acting on all the various
mass elements are
equivalent to a single
gravitational force
acting through a single
point called the center
of gravity (CG)
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…Center of Gravity
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The torque due to the gravitational force on an
object of mass M is the force Mg acting at the center
of gravity of the object
Problem-Solving Strategy –
Equilibrium Problems…
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Draw a diagram of the system
Isolate the object being analyzed
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If g is uniform over the object, then the center of
gravity of the object coincides with its center of mass
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If the object is homogeneous and symmetrical, the
center of gravity coincides with its geometric center
…Problem-Solving Strategy –
Equilibrium Problems…
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Establish a convenient coordinate system
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Find the components of the forces along the two
axes
…Problem-Solving Strategy –
Equilibrium Problems…
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Choose a convenient axis for calculating the
net torque on the object
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Apply the first condition for equilibrium (ΣF=0)
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Be careful of signs
Draw a free-body diagram
Show and label all external forces acting on the
object
Indicate the locations of all the forces
For systems with multiple objects, draw a
separate free-body diagram for each object
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Remember the choice of the axis is arbitrary
Choose an origin that simplifies the calculations
as much as possible
A force that acts along a line passing through the
origin produces a zero torque
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…Problem-Solving Strategy –
Equilibrium Problems
Weighted Hand Example…
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Apply the second condition for equilibrium (Στ = 0)
The two conditions of equilibrium will give a system
of equations
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Solve the equations simultaneously
If the solution gives a negative for a force, it is in the
opposite direction than what you drew in the free body
diagram
Model the forearm as a
rigid bar
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The weight of the
forearm is ignored
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There are no forces in
the x-direction
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Apply the first condition
for equilibrium (ΣFy = 0)
…Weighted Hand Example
Horizontal Beam Example…
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Apply the second
condition for equilibrium
using the joint O as the
axis of rotation (Στ =0)
Generate the equilibrium
conditions from the freebody diagram
Solve for the unknown
forces (F and R)
The beam is uniform
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So the center of gravity is
at the geometric center of
the beam
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The person is standing
on the beam
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What are the tension in
the cable and the force
exerted by the wall on
the beam?
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…Horizontal Beam Example...
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Draw a free-body
diagram
Use the pivot in the
problem (at the wall) as
the pivot
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…Horizontal Beam Example
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The forces can be
resolved into
components in the freebody diagram
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Apply the two
conditions of
equilibrium to obtain
three equations
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Solve for the unknowns
This will generally be
easiest
Note there are three
unknowns (T, R, θ)
Ladder Example…
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The ladder is uniform
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So the weight of the
ladder acts through its
geometric center (its
center of gravity)
…Ladder Example
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There is static friction
between the ladder and
the ground
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Draw a free-body diagram
for the ladder
The frictional force is ƒ = µn
Let O be the axis of rotation
Apply the equations for the
two conditions of
equilibrium
Solve the equations
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Ladder Example, Extended
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Add a person of mass M
at a distance d from the
base of the ladder
The higher the person
climbs, the larger the
angle at the base needs
to be
Eventually, the ladder
may slip
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