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!

3

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

TorquesDue to the weight

τg = –(L/2) mg sinθ1 (This torque is into the

page. Why?)

Due to the tension τT = LT sinθ2

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: μ n1 – n2 = 0Force, y: n1 – mg = 0Torque:

Ln2sinϕ – (L/2) mg cosϕ= 0

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Example – A Leaning Ladder

From the force equations we get n2 = μmg.

Therefore, μsinϕ – (1/2)cosϕ= 0

and so, tanϕ = 1/(2μ)

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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.