Transcript 12. Static Equilibrium - FSU Physics Department

12. Static Equilibrium
Conditions for Equilibrium
A bridge is an example
of a system in static
equilibrium. The bridge
undergoes neither
linear nor rotational
motion!
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Conditions for Equilibrium
A system is in static equilibrium if:
the net external force is zero
F  0
the net external torque is zero
  0
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Problem Solving Guideline
All static equilibrium problems are solved the same
way:
1. Find all external forces
2. Choose a pivot
3. Find all external torques
4. Set net force to zero
5. Set net torque to zero
6. Solve for unknown
quantities
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Problem Solving Guideline
It is generally simpler to choose the pivot at the
point of application of the force for which you
have the least information.
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Example – A Drawbridge
What is the tension in the
supporting cable of a 14 m,
11,000 kg drawbridge?
Forces:
1. Force at pivot
2. Tension in cable
3. Weight of bridge
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Example – A Drawbridge
Pivot:
A sensible choice is the
hinge since we do not
know the exact direction
of the hinge force, nor do
we care about it!
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Example – A Drawbridge
Torques:
Due to the weight
g = –(L/2) mg sinq1
(This torque is into the
page. Why?)
Due to the tension
T = LT sinq2
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Examples of
Static Equilibrium
Example – A Leaning Ladder
At what minimum angle can the ladder lean
without slipping?
The wall is frictionless and there
is friction between the floor and
the ladder.
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Example – A Leaning Ladder
Forces:
1. Normal force at bottom of ladder
2. Friction force at bottom of ladder
3. Ladder’s weight
4. Normal force at top of ladder
Pivot:
Choose bottom of ladder
Why?
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Example – A Leaning Ladder
Torques:
1. Due to ladder’s weight
2. Due to normal force at top of
ladder
Solve:
Force, x: m n1 – n2 = 0
Force, y: n1 – mg = 0
Torque:
Ln2sinf – (L/2) mg cosf = 0
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Example – A Leaning Ladder
From the force equations we get n2= m mg.
Therefore,
msinf – (1/2)cosf = 0
and so,
tanf = 1/(2m)
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Example – Standing on a Plank

F  0
  0
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Example – Standing on a Plank
Net force:
FL  FR  Mg  mg  0
Net torque:
 L  2d 
FR ( L  2d )  Mg 
  mgd  0
 2 
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Example – Standing on a Plank
d
1

m g
Force on right scale FR   M 
L  2d 
2
Ld 
1
m g
Force on left scale FL   M 
L  2d 
2
Do these make sense?
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Example – Force on Elbow
What is the force on the elbow?
m = 6 kg
Assume biceps force acts
3.4 cm from pivot point O.
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Example – Force on Elbow
Model forearm as a horizontal
rod

F  0
  0
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Example – Force on Elbow
The force we know least about is
the force on the elbow. So, let’s
take the elbow (O) as the pivot.
Net torque:
L
mh  Fm d  mgL  0
2
1
 L
Fm   mh  m  g
2
 d
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Example – Force on Elbow
Net force

F  0
x:
Fua, x  0  0  0  0
y:
Fua, y  Fm  mh g  mg  0
Fua, y  (m  mh ) g  Fm
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Summary

For a system to be in static equilibrium, both
the net force and the net torque must be zero.

When solving static equilibrium problems, it
often simplifies things to choose the pivot so
that the torque from unknown forces is zero.
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