1

Equilibrium

Objectives:

• Define and learn to solve problems in static equilibrium

• Define the 3 classes of levers and what each is best suited for

• Introduce stabilizing and dislocating forces

• Introduce dynamic equilibrium and the technique of inverse dynamics

Questions to Think About• Which of the following are better suited for moving

large loads? For moving loads quickly?

– Using your back extensors to lift a load at forearm’s length in front of you?

– Using your arms to perform a push-up?

• Why are the rotator cuff muscles important in preventing shoulder dislocations?

Static Equilibrium• A system is at rest and will remain at rest• No translation or rotation is occurring or will occur

• Conditions for static equilibrium(from Newton’s 1 st Law):

Σ T = 0

Σ Fx = 0

Σ Fy = 0

– Net external force in x direction equals zero

– Net external force in y direction equals zero

– Net torque produced by all external forces and all external torques equals zero

• Can use any point as the axis of rotation• Can solve for up to 3 unknown quantities

Example Problem #1During an isometric (static) knee extension, a

therapist measures a force of 100 N using a hand dynamometer in the position shown below

Find the resultant knee joint force and torque.

Fdyn = 100 N

m = 4.5 kg

60°

30 cm

KNEE

24 cm

2

Levers• Most skeletal muscles act using the principle of

leverage

• A lever system consists of:– An axis of rotation (or fulcrum)– A resistance force or load– An effort force

(the applied force that is used to move the load)

• There are 3 classes of leverFload

Feffort

axis of rotation

1st Class Lever• Effort force and load force are applied on opposite

sides of the axis of rotation

• Effort force and load force act in same direction

• For equilibrium: d⊥load Fload = d⊥effort Feffort

Fload Feffortd⊥load d⊥effort

axisFeffortFload

d⊥effortFload =d⊥load

Feffortor:

Mechanical Advantage

Feffort

Floadaxis

d⊥effortMechanical Advantage =d⊥load

• When Mechanical Advantage > 1:– Feffort needed is less than Fload

– Point at which Fload applied moves slower and shorter distance than point at which Feffort applied

– Good for strength, poor for moving load quickly or through large range of motion

Mechanical Advantage

Feffort

Fload

axis

• When Mechanical Advantage < 1:– Feffort needed is greater than Fload

– Point at which Fload applied moves faster and greater distance than point at which Feffort applied

– Good for moving load quickly or through large range of motion; poor for strength

• A 1st class lever can have a mechanical advantage greater than, equal to, or less than 1.

3

2nd Class Lever• Effort force and load force are applied on same

side of the axis of rotation• Effort force applied farther from axis than the load

force (i.e. d⊥effort > d⊥load)• Effort and load force act in opposite directions• Good for strength; poor for moving load quickly or

through large range of motion

Fload

Feffort

d⊥load

d⊥effort

axis

Feffort

Fload

3rd Class Lever• Effort force and load force are applied on same

side of the axis of rotation• Effort force applied closer to axis than the load

force (i.e. d⊥effort < d⊥load)• Effort and load force act in opposite directions• Good for moving load quickly or through large

range of motion; poor for strength

Fload

Feffort

d⊥loadd⊥effort

axis

Feffort

Fload

Stabilization vs. Dislocation• Muscles do not produce just torque; they also

produce stabilizing or dislocating forces at a joint.

• Can decompose a force into components parallel to(Fll) and perpendicular to (F⊥) the joint surface

• F⊥ points towards joint → stabilization

• F⊥ points away from joint → dislocation

FFll

F⊥

F

Fll

F⊥

Dislocation:Stabilization:

Joint Surface

The Last Example Problem!A person is holding her upper limb in the abducted

position shown.

Is the deltoid force stabilizing or dislocating?What class of lever is this?For equilibrium, is Fdeltoid greater than or less than FW?

Fdeltoid

FW = 35 N

30 cm

15 cm

Shoulder

30°

Join

t Sur

face

Hand

4

Dynamic Equilibrium• Applies to rigid bodies that are accelerating• Conditions for dynamic equilibrium

(from Newton’s 2 nd Law):

Σ T = I α

Σ Fx = m ax

Σ Fy = m ay

– Net external force in x direction equals mass times x accel.

– Net external force in y direction equals mass times y accel.

– Net torque produced by all external forces and torques equals moment of inertia times angular accel.

• Net torque typically computed about the center of mass• Can solve for up to 3 unknown quantities

Inverse Dynamics• To compute joint forces

and torques, body broken down into segments

• Solve equations of dynamic equilibrium for each segment

• Analyze from distal to proximal

FkneeX

FkneeY

FhipX

FhipY

FWthigh

Thip

Tknee

θthigh

FankleX

FankleY

FkneeX

FkneeY

FWleg

Tknee

Tankle

θleg

FankleY FWfoot

Tankle

θfoot

FankleX

FgrfY

FgrfX

Slow Movements• Analyzing forces acting under dynamic equilibrium

requires knowledge of accelerations

• If movements are slow: aX ≈ 0, aY ≈ 0, α ≈ 0

• In that case, can analyze using static equilibrium

• Simpler to understand, but will produce errors

θθαθ cos sin cos damdamIdFT YXW +−+=

θcos dFT W=

Dynamic (Fast)

Quasi-static (Slow)T m aX

aY

FW

α

dθ