Static Equilibrium. - Rice Universityphys101/ClassNotes/Static_Equilibrium.pdf · Static...

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PHYS101 Static Equilibrium – slide 1

PHYS101 - Static Equilibrium.

Dr. Suess

November 28, 2007

Static Equilibrium of a Rigid Body.

Static Equilibrium of aRigid Body.

• Static Equilibrium of aRigid Body.

• TranslationalEquilibrium

• Rotational Equilibrium

• Lines of Action (TwoForces)?

• Lines of Action (ThreeForces)?

• Problem Solving Tips- Static Equilibrium

• Problem Solving Tips- Static Equilibrium -cont’d

• The term equilibrium implies a body is at rest

• The term equilibrium implies a body is at rest or the center of

mass of the body moves with a constant velocity.

• We now wish to focus specifically on bodies that are at rest

(hence the term “static” equilibrium).

• Static equilibrium is of special interest to civil and mechanical

engineers.

engineers. Why?

◦ Structures such as bridges and buildings rely on a very sound

understanding of static equilibrium.

engineers. Why?

• So, what are the conditions which must be satisfied for a

complex, rigid body to be at static equilibrium under the action ofseveral forces?

engineers. Why?

◦ Consider translational equilibrium and rotational equilibrium

separately.

engineers. Why?

◦ Consider translational equilibrium and rotational equilibrium

separately. Remember, a rigid body can rotate even if the net

force on the body is zero!

Translational Equilibrium

• Translational equilibrium occurs when the linear acceleration of

the center-of-mass of the body is zero.

the center-of-mass of the body is zero. Think Newton’s 2nd law.

∑~Fext = M ~acm

∑~Fext = M ~acm = 0

• For translational equilibrium, the net force on the rigid body mustbe zero.

• This will ensure translational equilibrium.

• This will ensure translational equilibrium. NOT rotational

equilibrium !

Rotational Equilibrium

• Rotational equilibrium occurs when the angular acceleration of

the rigid body is zero.

the rigid body is zero. Think Newton’s 2nd law for rotation.

∑~τext = I ~α

∑~τext = I ~α = 0

• For rotational equilibrium, the net torque on a rigid body must bezero.

∑~τext = I ~α = 0

• This will ensure rotational equilibrium.

∑~τext = I ~α = 0

◦ For a rigid body to be in static equilibrium under the action of

two forces, the two forces must be of equal magnitude and

applied in opposite directions.

∑~τext = I ~α = 0

◦ For a rigid body to be in static equilibrium under the action of

two forces, the two forces must be of equal magnitude and

applied in opposite directions. In addition, the forces must

have the same line of action.

Lines of Action (Two Forces)?

For the rigid body shown above, the external forces applied to the

body are equal and opposite.

Lines of action are different.

The rigid object will rotate clockwise.

body are equal and opposite.

Lines of action are different.

The rigid object will rotate clockwise.

body are equal and opposite. The body will rotate clockwise about

its center of mass; therefore, the body is not in static equilibrium.

Now consider the case when the external forces applied to the

body are equal and opposite and lie on the same line of action.

Lines of action are the same.

The rigid object will NOT rotate clockwise.

Lines of action are the same.

The rigid object will NOT rotate clockwise.

The body will NOT rotate; therefore, the body is in static equilib-

Lines of Action (Three Forces)?

If three external forces applied to a body are concurrent (their lines

of action must intersect at a common point), the body can be in

static equilibrium as illustrated above.

Concurrent lines of action.

static equilibrium as illustrated above.

static equilibrium as illustrated above. To ensure the body is in

static equilibrium, the sum of the forces (and the torques) must

add up to zero.

add up to zero. This condition of concurrent lines of action is notmentioned in your textbook

add up to zero. This condition of concurrent lines of action is notmentioned in your textbook (wink-wink).

Problem Solving Tips - Static Equilibrium

• Identify the object in equilibrium.

• Identify all forces acting on the object, and

• Identify all forces acting on the object, and draw a

free-body-diagram showing the force vectors.

◦ It’s essential that the free-body-diagram shows the forces

drawn from their actual application points.

• Choose (and label) an appropriate coordinate system.

• Choose (and label) an appropriate coordinate system. Resolve

all vectors into components.

• Apply Newton’s 2nd law as a set of equations (one for each

component).

• Apply Newton’s 2nd law as a set of equations (one for each

component). Set each equation to zero.

Problem Solving Tips - Static Equilibrium - cont’d

• Choose a pivot point for calculating torques.

• Choose a pivot point for calculating torques. The choice is yours,

but a well-thought out choice will simplify the algebra.

• Calculate the torque (~τ =∑

i(~ri ×

~Fi)) associated with each

force and sum the torques to zero.

i(~ri ×

• Solve the system of force and torque equations for the unknowns

of the problem.

i(~ri ×

• Solve the system of force and torque equations for the unknowns

of the problem.

• Sit back and relax - knowing that your efforts may have stabilized

a bridge or the building you helped design will remain static.

Static Equilibrium. - Rice Universityphys101/ClassNotes/Static_Equilibrium.pdf · Static...

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