Download Static Equilibrium

Total Equilibrium
In general, there are six degrees of freedom (right, left,
up, down, ccw, and cw):
ccw (+)
cw (-)
Fx= 0
Right = left
Fx= 0
Up = down
(ccw)=
(ccw)
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 (and probably no one else!)
Conditions for Equilibrium
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The net force equals zero
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F=0
If the object is modeled as a particle, then
this is the only condition that must be
satisfied
The net torque equals zero
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=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,
=I
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For rotational equilibrium,
=0
This must be true for any axis of
rotation
Equilibrium Summary
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There are two necessary conditions for
equilibrium
The resultant external force must equal zero:
F=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
The resultant external torque about any axis
must be zero:
=0
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This is a statement of rotational equilibrium
The angular acceleration must equal zero
Static vs. Dynamic Equilibrium
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In this chapter, we will concentrate on
static equilibrium
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Dynamic equilibrium is also possible
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The object will not be moving
The object would be rotating with a
constant angular velocity
In either case, the
=0
Equilibrium Equations
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We will restrict the applications to
situations in which all the forces lie in
the xy plane
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These are called coplanar forces since they
lie in the same plane
There are three resulting equations
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Fx = 0
Fy = 0
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)
Center of Gravity, cont
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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
If g is uniform over the object, then the
center of gravity of the object coincides with
its center of mass
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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Draw a diagram of the system
Isolate the object being analyzed
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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
Problem-Solving Strategy –
Equilibrium Problems, 2
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Establish a convenient coordinate
system
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Find the components of the forces along
the two axes
Apply the first condition for equilibrium
( F=0)
Be careful of signs
Problem-Solving Strategy –
Equilibrium Problems, 3
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Choose a convenient axis for calculating
the net torque on the 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
Problem-Solving Strategy –
Equilibrium Problems, 4
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Apply the second condition for
equilibrium ( = 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
Weighted Hand Example
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Model the forearm
as a rigid bar
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The weight of the
forearm is ignored
There are no forces
in the x-direction
Apply the first
condition for
equilibrium ( Fy =
0)
Weighted Hand Example, cont
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Apply the second
condition for equilibrium
using the joint O as the
axis of rotation ( =0)
Generate the
equilibrium conditions
from the free-body
diagram
Solve for the unknown
forces (F and R)
Horizontal Beam Example
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The beam is uniform
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So the center of gravity
is at the geometric center
of the beam
The person is standing
on the beam
What are the tension in
the cable and the force
exerted by the wall on
the beam?
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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This will generally be
easiest
Note there are three
unknowns (T, R, )
Horizontal Beam Example
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The forces can be
resolved into
components in the
free-body diagram
Apply the two
conditions of
equilibrium to obtain
three equations
Solve for the
unknowns
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)
There is static
friction between the
ladder and the
ground
Ladder Example
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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
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
General Procedure:
• Draw free-body diagram and label.
• Choose axis of rotation at point where least
information is given.
• Extend line of action for forces, find moment
arms, and sum torques about chosen axis:
• Sum forces and set to zero:
• Solve for unknowns.
Fx= 0;
Fy= 0
Example 1: Find the tension in
the rope and the force by the
wall on the boom. The 10-m
boom weighing 200 N. Rope is
2 m from right end.
T
300
800 N
For purposes of summing torques, we consider
entire weight to act at center of board.
Fy
T
Fx
300
200 N
T
5m
800 N
300
3m
200 N
2m
800 N
T
Example 1
(Cont.)
300
200 N
Fy
r
Fx
5m
800 N
T
300
3m
2m
200 N
800 N
Choose axis of rotation at wall (least information)
(ccw):
Tr = T (8 m)sin 300 = (4 m)T
(cw):
(200 N)(5 m) + (800 N)(10 m) = 9000 Nm
(4 m)T = 9000
Nm
T = 2250 N
T
Example 1
(Cont.)
Fy
300
200 N
T
Fx Tx
5m
Ty
30300 0
3m
2m
200 N
800 N
800 N
F(up) = F(down): Ty + Fy = 200 N + 800 N
Fy = 200 N + 800 N - Ty ;
Fy = 1000 N - T sin 300
Fy = 1000 N - (2250 N)sin 300
Fx = Ty = (2250 N) cos 300
F(right) = F(left):
Fx = 1950 N
Fy = -125 N
or
F = 1954 N, 356.30
Summary: Procedure
• Draw free-body diagram and label.
• Choose axis of rotation at point where least
information is given.
• Extend line of action for forces, find moment
arms, and sum torques about chosen axis:
• Sum forces and set to zero:
• Solve for unknowns.
Fx= 0;
Fy= 0