lecture note statics equilibrium
DESCRIPTION
Statics Notes for engineeringTRANSCRIPT
12. Static Equilibrium
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Conditions for Equilibrium
A bridge is an example of a system in static equilibrium. The bridgeundergoes neitherlinear nor rotationalmotion!
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0
the net external torque is zero
0F
the net external force is zero
Conditions for Equilibrium
A system is in static equilibrium if:
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Problem Solving Guideline
All static equilibrium problems are solved the same way:
1. Find all external forces2. Choose a pivot3. Find all external torques4. Set net force to zero5. Set net torque to zero6. 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 thesupporting cable of a 14 m,11,000 kg drawbridge? Forces:
1. Force at pivot2. Tension in cable3. Weight of bridge
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Example – A Drawbridge
Pivot:A sensible choice is the hinge since we do notknow 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 sin1
(This torque is into the page. Why?)
Due to the tension T = LT sin2
Examples of Static Equilibrium
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Example – A Leaning Ladder
At what minimum angle can the ladder leanwithout 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 ladder2. Friction force at bottom of ladder3. Ladder’s weight4. 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 weight2. Due to normal force at top ofladder
Solve:Force, x: n1 – n2 = 0Force, y: n1 – mg = 0Torque:
Ln2sin – (L/2) mg cos= 0
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Example – A Leaning Ladder
From the force equations we get n2= mg.
Therefore, sin – (1/2)cos= 0
and so, tan = 1/(2)
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Example – Standing on a Plank
0
0F
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Example – Standing on a Plank
0RL M mgF gF
2( 2 ) 0
2R
L dL d Mg mF gd
Net force:
Net torque:
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Example – Standing on a Plank
1
2 2R
dM m g
L dF
Force on right scale
Do these make sense?
1
2 2L
L dM m g
L dF
Force on left scale
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Example – Force on Elbow
What is the force on the elbow?
m = 6 kg
Assume biceps force acts3.4 cm from pivot point O.
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Example – Force on Elbow
0
0F
Model forearm as a horizontal rod
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Example – Force on Elbow
02h m
Lm F d mgL
The force we know least about is the force on the elbow. So, let’stake the elbow (O) as the pivot.
Net torque:
1
2m h
LF m m g
d
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Example – Force on Elbow
, 0 0 0 0ua xF
Net force
, 0mua y hF F m g mg
0F
x:
y:
, ( )hua y mm mF g F
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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.