CHAPTER 12 : STATIC EQUILIBRIUM AND ELASTICITY 12.1) The Conditions For Equilibrium

Object treated as a particle • One necessary condition for equilibrium = the net force acting on an object is zero.

For extended object • The net force acting on an object is zero.

• Involves the net torque acting on the extended object.

• Equilibrium does not require the absence of motion.

• A rotating object can have constant angular velocity and still be in equilibrium.

• Consider a single force (

Figure (12.1)

).

F

acting on a rigid object • The effect of the force depends on its point of application P.

• If

r

is the position vector of this point relative to O, the torque assiciated with the force (11.7) :

F

about O is given by Equation

τ



r



F

• From vector product (Section (11.2)) – the vector perpendicular to the plane formed by

r

and

F

.

 is • Right-hand rule – to determine the direction of  : Curl the fingers of your right hand in the direction of rotation that

F

tends to cause about an axis through O, your thumb then points in the direction of  .

• In Figure (12.1)  is directed toward out of the page.

• From

Figure (12.1)

– the tendency of

F

to rotate the object about an axis through O depends on the moment arm d and the magnitude of

F

.

• Magnitude of  = Fd (Eq. (10.19).

• Suppose a rigid object is acted on first by force later by force

F

2 .

F

1 and • If the two forces have the same magnitude – they will produce the same effect on the object only if they have the same direction and the same line of action.

• Equivalent forces = two forces F 1 and F 2

are equivalent if and only if

F 1 = F 2

and if and only if the two produce the same torque about any axis.

•

Figure (12.2)

– two forces are equal in magnitude and opposite in direction = not equivalent.

F

2 Figure (12.2)

F

1 O • The force directed to the right tends to rotate the object clockwise about an axis perpendicular to the diagram through O.

• The force dircted to the left tends to rotate it counterclockwise about that axis.

•

Figure (12.3)

– an object is pivoted about an axis through its center of mass.

• Two forces of equal magnitude act in opposite directions along parallel lines of action.

• A pair of forces acting in this manner = couple • Because each force produces the same torque Fd, the net torque has a magnitude of 2Fd.

• • The object rotates clockwise and undergoes an angular acceleration about the axis = nonequilibrium situation (with respect to rotational motion).

The net torque on the object gives rise to an angular acceleration  according to the relationship  = 2Fd = I  (Eq. (10.21)).

• • An object is in rotational equilibrium only if its angular acceleration  = 0.

Because  = I  for rotation about a fixed axis, the net torque about any axis must be zero.

Two necessary conditions for equilibrium of an object : 1.

The resultant external force must equal zero.



F

= 0 (12.1) Translational equilibrium – tells us that the linear acceleration of the center of mass of the object must be zero when viewed from an inertial reference frame.

2.

The resultant external torque about any axis must be zero.

 = 0 (12.2) Rotational equilibrium and tells us that the angular acceleration about any axis must be zero.

3.

Special case of static equilibrium – the object is at rest and so has no linear or angular speed (that is, v  = 0).

CM = 0 and

• From equation (12.1) and (12.2) – equivalent to six scalar equations (F x , F y , F z ,  x ,  y ,  z ).

• Restrict to situations in which all the forces lie in the xy plane.

• Forces whose vector representations are in the same plane are said to be coplanar.

• Deal with only three scalar equations.

• Two of these come from balancing the forces in the x and y direction (F x , F y ).

• The third comes from the torque equation – the net torque about any point in the xy plane must be zero.

• Hence, the two conditions of equilibrium provde the equations :  F x = 0  F y = 0  z = 0 (12.3) where the axis of the torque equation is arbitrary.

• Regardless of the number of forces that are acting – if an object is in translational equilibrium and if the net torque is zero about one axis, then the net torque must also be zero about any other axis.

• The point can be inside or outside the boundaries of the object.

• • Consider an object being acted on by several forces such that the resultant force 

F

=

F

1 +

F

2 +

F

3 + … = 0.

Figure (12.4)

– four forces acted on the object.

F

1

F

2 O

r

1

r

’

r

1

O’ –

r

’

Figure (12.4)

F

3

F

4 • The point of application of the position vector

r

1 .

F

1 relative to O is specified by • Similarly, the points of application of specified by

r

2 ,

r

3 , ….

F

2 ,

F

3 , … are • The net torque about an axis through O is : 

τ

O 

r

1 

F

1 

r

2 

F

2 

r

3 

F

3  ...

• Now consider another arbitrary point O’ having a posiiton vector

r

’ relative to O.

• The point of application of the vector

r

1 –

r

’.

F

1 relative to O’ is identified by • The point of application of

F

2 so forth.

relative to O’ is

r

2 –

r

’ , and • Therefore, the torque about an axis through O’ is : 

τ

O '  

(

r

1

r

1  

F

1

r

' )

 

r

2

F

1  

F

2

(

r

2  

r

3

r

' )

 

F

3

F

2  

...



(

r

3

r

'

 

r

' (

F

1

)

 

F F

2 3 

...



F

3 

...)

• Because the net force is assumed to be zero (given that the object is in translational equilibrium), the last term vanishes, and we see that the torque about O’ is equal to the torque about O.

• Hence, if an object is in translational equilibrium and the net torque is zero about one point, then the net torque must be zero about any other point.

12.2) More on the Center of Gravity

• Whenever deal with a rigid object – consider (1) the force of gravity acting on it, and (2) the point of application of this force.

• On every object is a special point = center of gravity.

• All the various gravitational forces acting on all the various mass elements of the object are equivalent to a single gravitational force acting through this point.

• To compute the torque due to the gravitational force on an object of mass M – consider the force M

g

acting at the center of gravity of the object.

To find this special point (center of gravity) • If we assume that

g

is uniform over the object – the center of gravity of the object coincides with its center of mass.

• Consider an object of arbitrary shape lying in the xy plane (

Figure (12.5)

).

• Suppose the object is divided into a large number of particles of masses m (x 1 , y 1 ), (x 2 , y 2 ), (x 3 1 , y 3 , m 2 ), ….

, m 3 , … having coordinates • In Equation (9.28) – the x coordinate of the center of mass of such an object :

x

CM 

m

1

x

1 

m

1

m

 2

x m

2 2  

m m

3 3

x

 3 

...

...

 

m

n

m

n

x

n   i  i

m

i

x

i

m

i Similar form of equation to define the y coordinate of the center of mass.

•

Figure (12.6)

– consider the force of gravity exerted on each particle.

• Each particle contributes a torque about the origin equal in magnitude to the particle’s weight mg multiplied by its moment arm.

• Example – the torque due to the force m 1

g

1 is m 1 g 1 x 1 , where g 1 is the magnitude of the gravitational field at the position of the particle of mass m 1 .

• Locate the center of gravity, the point at which application of the single gravitational force M

g

(where M = m 1 + m 2 + m 3 + … is the total mass of the object) has the same effect on rotation as does the combined effect of all the individual gravitational forces m i

g

i .

• Equating the torque resulting from M

g

acting at the center of gravity to the sum of the torques acting on the individual particles gives :  m 1 g 1  m 2 g 2  m 3 g 3  ...

 x CG   m 1 g 1 x 1 m 3 g 3 x 3  m  ...

2 g 2 x 2 • If we assume uniform g over the object – then the g terms cancel and we obtain :

x

CG 

m

1

x

1

m

1  

m

2

x m

2 2  

m m

3 3 

x

3

...



...

(12.4) The center of gravity is located at the center of mass as long as the object is in a uniform gravitational field.

12.4) Elastic Properties of Solids

• So far, we assume that objects remain undeformed when external forces act on them.

• In reality – all objects are deformable.

• That is, it is possible to change the shape or the size of an object (or both) by applying external forces.

• As these changes take place, however, internal forces in the object resist the deformation.

• The deformation of solids is discussed in terms of two concept: (1) stress, and (2) strain.

•

Stress =

quantity that is proportional to the force causing a deformation; more specifically stress is the external force acting on an object per unit cross-sectional area.

•

Strain =

a measure of the degree of deformation.

• For sufficiently small stresses, strain is proportional to stress; the constant of proportionality depends on the material being deformed and on the nature of the deformation.

• The proportionality constant is called the elastic modulus.

• The elastic modulus is therefore the ratio of the stress to the resulting strain :

Elastic modulus



stress strain

(12.5) A comparison of what is done (force applied) to a solid object and how that object responds (deforms)

• Consider three types of deformation and define an elastic modulus for each : 1)

Young’s modulus

change in its length.

– measures the resistance of a solid to a 2)

Shear modulus

– measures the resistance to motion of the planes of a solid sliding past each other.

3)

Bulk modulus

– measures the resistance of solids or liquids to changes in their volume.

Young’s Modulus : Elasticity in Length

• Consider a long bar of cross-sectional area A and initial length L i that is clamped at one end (

Figure (12.13)

).

Figure (12.13) • When an external force is applied perpendicular to the cross section, internal (“stretching”).

forces in the bar resist distortion • But the bar attains an equilibrium in which its length L f is greater than L i and in which the external force is exactly balanced by internal forces.

• In such situation, the bar is said to be stressed.

• We define the

tensile stress

as the ratio of the magnitude of the external force F to the cross-sectional area A.

• The

tensile stress

change in length in this case is defined as the ratio of the  L to the original length L i .

• We define ratios :

Young’s modulus

by a combination of these two

Y



tensile stress tensile strain



F / A



L / L

i (12.6) Young’s modulus • Young’s modulus – used to characterize a rod or wire stressed under either tension or compression.

• Because strain is a dimensionless quantity, Y has units of force per unit area.

• Experiments show : (a) that for a fixed applied force, the change in length is proportional to the original length and (b) that the force necessary to produce a given strain is proportional to the cross-sectional area.

• The

elastic limit

of a substance is defined as the maximum stress that can be applied to the substance before it becomes permanently deformed.

• It is possible to exceed the elastic limit of a substance by applying a sufficiently large stress (Figure (12.14)).

• When the stress exceeds the elastic limit, the object is permanently distorted and does not return to its original shape after the stress is removed.

• The shape of the object is permanently changed.

• As the stress is increased even further, the material ultimately breaks.

Shear Modulus : Elasticity of Shape

•

Figure (12.15)

– another type of deformation occurs when an object is subjected to a force tangential to one of its faces while the opposite face is held fixed by another force.

• The stress in this case is called a shear stress.

Figure (12.15) • If the object is originally a rectangular block, a shear stress results in a shape whose cross-section is a parallelogram.

• A book pushed sideways (Figure (12.15b) is an example of an object subjected to a shear stress.

• For small distortions, no change in volume occurs with this deformation.

• The

shear stress

is defined as F/A, the ratio of the tangential force to the area A of the face being sheared.

• The

shear strain

is defined as the ratio  x / h, where  x is the horizontal distance that the sheared face moves and h is the height of the object.

• The

shear modulus

is :

S



shear stress shear strain



F /



x / A h

(12.7) Shear modulus The unit of shear modulus is force per unit area.

Bulk Modulus : Volume Elasticity

• Bulk modulus characterizes the response of a substance to uniform squeezing or to a reduction in pressure when the object is placed in a partial vacuum.

• Suppose that the external forces acting on an object are at right angles to all its faces (

Figure (12.16)

), and that they are distributed uniformly over all the faces.

Figure (12.16) • Such a uniform distrivution of forces occures when an object is immersed in a fluid (Chapter 15).

• An object subject to this type of deformation undergoes a change in volume but no change is shape.

• The

volume stress

is defined as the ratio of the magnitude of the normal force F to the area A.

• The quantity P = F / A is called the

pressure

.

• If the pressure on an object changes by an amount  P =  F / A, then the object will experience a volum change  V.

• The

volume strain

is equal to the change in volume divided by the initial volume V i .

 V • From Equation (12.5) compression in terms of the as :

B



volume volume stress strain

characterize a volum (“bulk”)

bulk modulus

, which is defined   

F /



V / A V

i    

V / P V

i (12.8) Bulk modulus

• A negative sign is inserted in this defining equation so that B is a positive number.

• An increase in pressure (positive volume (negative  P) causes a decrease in  V) and vice versa.

Example (12.7) : Squeezing a Brass Sphere

A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0 x 10 5 N/m 2 (normal atmospheric pressure). The sphere is lowered into the ocean to a depth at which the pressure is 2.0 x 10 7 N/m 2 . The volume of the sphere in air is 0.50 m 3 . By how much does this volume change once the sphere is submerged?