Topic 2.2 ExtendedF – Torque, equilibrium, and stability

TORQUE

Torque τ is the rotational equivalent of force.Consider a door, as viewed from above.

WALL WALL

The location of the force will determine the ease with which the door opens.

F0 F1F2

r1

r2

Easier HarderF0F1F2

It is clear that the turning force τ is proportional to the distance r from the pivot point.

r2 > r1 > r0

If we now choose one r, the angle θ that F makes with r will also determine the ease with which the door opens.

θ1θ2 θ3

Easier Harder

θ1 θ2 θ3> >

The turning force τ is dependent on the angle between the force and the position vector.

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

TORQUE

WALL WALL

In fact, if θ is zero, so is the turning force τ.Finally, the turning force τ is proportional to the size of the force.Putting it all together we have

τ = rF sin θ Definition of torque τ

Question: Why do we use sine instead of cosine?Suppose we apply a force of 80 n a distance of 25 cm from the hinge, at an angle of 30° with respect to r. Then the torque is

0.25 m

80 n30°

τ = rF sin θ= (0.25 m)(80 n) sin 30°= 10 n·m

FYI: Torque has the same units as a force times a distance (work). But torque is not work and so is never written as joules (J).

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

TORQUEConsider a disk that can rotate about its center.

Suppose we apply an identical force F to the disk at four different points.

F

F

F

F

A

B

CF

D

Although F is the same at each point, r is in a different direction:

rr

r

r

FYI: We call the line PARALLEL to the force the force line of action.

line of action

FYI: We call the line PERPENDICULAR to the force the lever arm or the moment arm.

moment or lever arm

The length of the lever arm is r sin so that if we want, we can rewrite our torque equation like this:

τ = rF sin θ

τ = F(r sin θ)Thus"The torque is the force times the lever arm."

FYI: Even though all of the forces are identical, each force produces a different rotation.

Which force produces the greatest torque?

Which force produces the least torque?

Which force/s produce/s a ccw torque?

Which force/s produce/s a cw torque?

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

EQUILIBRIUMConsider a disk that can rotate about its center.Suppose we apply two forces of equal value F to the disk at the two points shown.

F

F

A

B

The force at A wants to turn the wheel ccw. The force at B wants to turn the disk cw. Thus

= 0 Condition for Rotational Equilibrium

FYI: The disk is in rotational equilibrium but the two forces still cause the disk to have a net force up and to the left. It is NOT in translational equilibrium.

FYI: If an object is in rotational equilibrium then its angular speed will NOT change. If it is not spinning it will continue to not spin. If it is spinning at a particular angular speed, it will continue to spin at that speed.

If we reverse the force at B the conditional for rotational equilibrium is not satisfied since both forces now want to produce a ccw torque:

F

The force at A and the force at B now cancel, though, so that the center of mass does not accelerate. Thus

F = 0 Condition for Translational Equilibrium

FYI: A body is said to be in MECHANICAL EQUILIBRIUM when the conditions for BOTH rotational and translational equilibrium are satisfied.

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

EQUILIBRIUMSuppose a uniform beam of mass m and length L is placed on two scales, as shown.

It is expected that each scale will read the same, namely half the weight of the beam.Now we place a block of mass M on the beam, closer to the left-hand scale.

M

It is expected that the left scale will read higher than the right one, because the block is closer to it.

X

FYI: We say that an object is in static equilibrium when it is in mechanical equilibrium AND it is not moving. Both of these situations are examples of static equilibrium.

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

EQUILIBRIUMTo analyze an extended system we use what we will call an extended free-body diagram.

Find N1 and N2 in terms of x, L, m, and M and g.

mg

L/2

N1 N2Mg

x

FYI: We must use an extended diagram, since there are torques present.

From our balance of forces we have

ΣFy = 0

N1 + N2 – Mg – mg = 0

Note: We have ONE equation with TWO unknowns. We must use the balance of torques to obtain the other equation.

EQUILIBRIUM

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

Find N1 and N2 in terms of x, L, m, and M and g.

mg

L/2

N1 N2Mg

x

In order to use our balance of torques we need to choose a pivot point.

FYI: If a system is in static equilibrium you can use ANY point as the pivot.

I have arbitrarily chosen the left end of the board as my pivot point.

+

I have arbitrarily chosen CLOCKWISE to be a positive torque.

NOTE: Mg and mg want to rotate the lever arm in the POSITIVE direction, and N2 provides the balancing NEGATIVE torque.

Στ = 0

0·N1 + x·Mg + (L/2)mg – L·N2 = 0

FYI: Choosing the pivot point at the application point of a force eliminates that force from the balance of torques equation.

EQUILIBRIUM

Rotational Motion and Equilibrium8-2 Torque, Equilibrium and Stability

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

Find N1 and N2 in terms of x, L, m, and M and g.

mg

L/2

N1 N2Mg

x

We now resolve our system of equations:

+

0·N1 + x·Mg + (L/2)mg – L·N2 = 0

N1 + N2 – Mg – mg = 0

N2 = + gxML

m2

N1 = (M + m)g - N2

Note: If x = L/2, N2 reduces to the expected (M + m)g/2 (exactly half the total weight). N1 will reduce to exactly the same thing.

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

Find N1 and N2 if x = 2 m, L = 20 m, the mass of the plank is 50 kg, and the mass of the block is 100 kg.We have

N2 = + gxML

m2

= + ·102·10020

502

= 350 n

N1 = (M + m)g - N2 = (100 + 50)10 - 350 = 1150 n

Question: Why is N1 > N2?

EQUILIBRIUM

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

In the previous example all of the torques were caused by perpendicular forces, so no angles were used (since sin 90° = 1).Now consider the boom crane shown here:The critical components of the crane need to be designed to withstand the forces (and torques) the crane is subjected to.We need to know what tension the support cables must withstand.We need to know what forces the support pin must withstand.

Improper structural analysis may result in DEATH!

WARNING

EQUILIBRIUM

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

Let’s do some GOOD analysis!Here are the variables: θ

m

T

FH

FV

mg

Mg

M

Our goal is to find the three green forces, and the net force on the pin.An extended free-body diagram simplifies our analysis.

T

FH

FV

mg

Mg

L/2x θ

If L is the length of the beam, L/2 is the effective position of the beam’s weight Mg.

Let x be the distance mg is from the pin.

EQUILIBRIUM

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

Suppose θ = 30°, the length of the beam is 20 m, and the mass of the beam is 400 kg.Then our diagram reduces to…

T

FH

FV

mg

4000 n

10x 30°

60°

60°

ΣFx = 0T - FH = 0

T = FH

ΣFy = 0FV – 4000 - mg = 0

FV = 4000 + mg

Στ = 0

10·4000 sin 60° + xmg sin 60° - 20·T sin 30° = 0

x

T = 3464.1 + .866xm

EQUILIBRIUM

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

Our three unknowns in terms of x and m are thus:

T

FH

FV

mg

4000 n

10x 30°

60°

60°

FV = 4000 + 10m

x

T = FH = 3464.1 + .866xm

As a specific example, if m = 2000 kg, and x = 15 m, then…

T = FH = 3464.1 + .866·15·2000 = 29445 n

FV = 4000 + 10·2000 = 24000 n

FPIN2 = FV

2 + FH2

To find the net force on the pin…

FPIN2 = 240002 + 294452

FPIN = 37987 n

EQUILIBRIUM

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

EQUILIBRIUM

Consider a typical building shown hereVertical supports called columns transfer the weight of the building to the ground.Horizontal supports called spans transfer the weight of the floors to the columns.The building is in static equilibrium.If explosive charges are detonated at the base of the columns, the building is no longer in equilibrium.

FYI: Of course, after the collapse the debris reestablish static equilibrium conditions.

FYI: The fun begins when we wish to remove the conditions of static equilibrium, as in this example...

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

STABILITY

Before we talk about stability we have to define three types of equilibrium.Consider two bowls and a level table top.If we carefully place a ball on each surface it will be in equilibrium:

STABLE NEUTRAL UNSTABLE

EQUILIBRIUM EQUILIBRIUM EQUILIBRIUM

Now we gently displace each ball to the right and observe the results...

FYI: The STABLE configuration has a built-in RESTORING FORCE that tries to return the mass to its equilibrium position.

FYI: The UNSTABLE configuration has a built-in ANTI-RESTORING FORCE that tries to remove the mass from its equilibrium position.

FYI: The NEUTRAL configuration has neither built-in force.

Topic 2.2 ExtendedF – Torque, equilibrium, and stability

STABILITY

Now we can look at extended solids.Consider the three solids, all of which are in equilibrium.

cg cgcg

FYI: As long as the center of gravity is DIRECTLY above the BASE OF CONTACT, the extended object is in equilibrium.

mg mgmg

mg

FYI: mg provides a restoring torgue so that this body is in STABLE EQUILIBRIUM.

cg

mg

FYI: mg provides NO torque so that this body is in NEUTRAL EQUILIBRIUM.

mg

FYI: mg provides an anti-restoring torque so that this body is in UNSTABLE EQUILIBRIUM.

FYI: If we rotate the stable extended body far enough, it will become unstable:

mg mgmg