14
Static Chapter Equilibrium II
Chapter 15
Oscillatory Motion
Conditions for Equilibrium
A body is in static equilibrium when it is stationary and when both "the net external force and the net external torque are zero!"
Fi 0
i
and
i 0
i
If the tension in the support cable does not act at the center of "gravity" for the beam and the worker then the beam will rotate! Not good! For static equilibrium we also have
pi 0
i
net R cm M g
Example: Teeter-Totter
Consider a teeter-totter as shown in the figure. If a child of mass m is a distance x from the fulcrum and the weight of mass M is a distance y, what is the ratio of x/y for the teeter-totter to be in equilibrium? As usual we must first choose an origin from which to find the torques. First we consider the fulcrum as the origin:
mgx Mgy x/y M/m
The larger the mass M, the further the child must sit from the fulcrum. Now choose the location of the weight as the origin. The normal force at the fulcrum must equal the weight of the child and the mass M. Hence:
mgx y M m gy mx My
This leads to the same result. What if the board is at an angle q wrt the horizontal?
Example: Crane Equilibrium
Consider the crane as shown in the figure. The boom is supported by a cable attached to its center point. Find the tension in the cable when angle of the boom is 50o above the horizontal and the mass of the boom is 1700kg. The hanging mass is 2200kg.
Summing torques about the pivot point P:
T9 sin50 - 1700g9 cos 50 - 2200g18 cos 50 0
Solving for T:
T 4400 1700 9. 8 cos 50 50kN sin50
Again the normal forces at the pivot balance all of the forces.
Example: Tip or Slide, that is the Question.
A rectangular block, twice as high as it is wide, is resting on an incline with an angle q and a coefficient of static friction m. What condition on m will cause the block to slide before it tips? The maximum torque resulting from the normal force occurs when the normal force acts at the lowest edge of the block. If the torque from the gravitational force is greater than this restoring torque, the block will tip! The torque about the lowest edge must vanish. This means that at the steepest angle prior to tipping the center of mass must lie directly above this point. This angle is given by tan q = . For the block not to slide:
mg cos - mg sin 0 tan 1/2
Hence if m is less than the block will slide before it tips!
Example: Tip or Slide, that is the Question.
A cone, three times as high as it is wide, is resting on an incline with an angle q and a coefficient of static friction m. What condition on m will cause the cone to slide before it tips?
Again the torque about the lowest edge must vanish. For a cone the center of mass is h/4 above the base. At steepest angle prior to tipping the center of mass must lie directly above the lowest edge. This angle is given by tan q = 2/3. For the block not to slide:
mg cos - mg sin 0 tan 2/3
Hence if m is less than 2/3 the block will slide before it tips!
Example: Weighted Disk on Incline
Consider a uniform disk of mass M and radius R with a weight of mass m on the rim. The disk is on an incline of angle q with friction (rolls and not slides). Find the angle f as shown such that the disk is in static equilibrium. In equilibrium the forces are balanced by the normal force and the frictional force. To balance torques it is convenient to choose the point of contact between the incline and the disk as the origin. The moment arm to the center of the disk is: The moment arm for the weight on the rim is: The expression for the total torque:
R R sin i R cos j
r R - R cos i R sin j
R -Mg j r -mg j 0 -MgR sin k - mgRsin - cos k 0
Example: Weighted Disk on Incline
Consider a uniform disk of mass M and radius R with a weight of mass m on the rim. The disk is on an incline of angle q with friction (rolls and not slides). Find the angle f as shown such that the disk is in static equilibrium. The expression for the net torque:
-MgR sin k - mgRsin - cos k 0
Solving for f :
m cos m sin M sin cos 1 M/m sin cos -11 M/m sin
The normal force is what is required for equilibrium normal to the incline. The friction force results in equilibrium parallel to the incline. It is convenient to choose the origin for the moment arms at the point of contact between incline the and the disk so that it is not required to know these forces.
Example: Equilibrium Conditions
The potential energy as a function of x is:
Ux U o x3 a x2 4 x xo x3 x2 o o
For what values of a will there be two static equilibria? Comment on the stability of these equilibria. Equilibrium is determined by dU/dx = 0. Taking the derivative of U:
dU U o xo dx
x 2 2a x 4 3 2 xo xo
x 0 x1,2 1 -a a 2 - 12 o 3
For two real roots a2 > 12, a ~ 3.5. Taking the second derivative yields:
d 2Ux 1,2 o U2 2 dx xo x 6 x1,2 2a o
o 2 U2 xo
a 2 - 12
Right point is metastable. Left point is unstable.
Example: Unstable Equilibrium
The top of a roller coaster track is described by:
h . 94x -. 01x 2
Here h and x are measured in meters. (a) Find the point on the track where the cars are in static equilibrium. (b) Is the equilibrium stable or unstable? (c) What is the height of the track at the equilibrium point? (a) The potential energy is given by U(x) = mgh. For equilibrium dU/dx = 0.
dU mg. 94 -. 02x 0 x 47m dx (b) The second derivative determines stability:
d 2U -. 02mg 0 unstable dx 2
(c) The height at x = 47 is:
h . 9447 -. 0147 2 22m
Example: Multiple Equilibrium Points
In a certain semiconductor the local potential is:
Ux ax 2 - bx 4 Here x is the position of the electron in nm, and U is its potential energy in aJ = 10-18J. The constants are a = 8 aJ/nm2 and b = 1 aJ/nm3. Locate the equilibrium positions for the electron and describe their stability.
For equilibrium dU/dx = 0:
dU 2ax - 4bx 3 0 dx x 0, a/2b 0, 2nm
The second derivative determines stability:
d 2Ux 2a - 12bx 2 2 dx d 2Ux 0 2a 16 0 stable dx 2 d 2Ux 2 16 - 124 0 unstable 2 dx
Chapter 15
Oscillatory Motion
Simple Harmonic Motion
Simple harmonic motion results when an object is subject to a linear restoring force and is called simple harmonic motion, SHO.
Mathematically such a force is described as:
F -kx
The is the force exerted by an ideal spring of spring constant k. From Newton's 2nd we can write:
d 2x -kx Fm 2 dt
An object experiencing such a force means that when it is displaced from equilibrium there is a force proportional to the distance from equilibrium that accelerates the object back towards its equilibrium position. How do we describe such motion?
Simple Harmonic Motion
Simple harmonic motion results when an object is subject to a linear restoring force and is called simple harmonic motion, SHO.
From Newton's 2nd:
d 2x -kx Fm 2 dt
The solution to this equation is of the form x = A cos(wt) + B sin(wt). To see this we simply substitute this function into the "differential equation" represented by Newton's 2nd. Taking the derivatives:
dx -Asint B cos t dt d 2x - 2Acos t - 2B sint dt 2 d 2x - 2x dt 2
Simple Harmonic Motion
Simple harmonic motion results when an object is subject to a linear restoring force and is called simple harmonic motion, SHO.
Substituting this result into Newton's
2nd:
d 2x -m 2x -kx Fm 2 dt
This equation is satisfied if w2 = k/m ! But what is this w ? We know from the form of x that when wT = 2p the sine or cosine function returns to its value it had when t = 0. So T is the period of the sine or cosine function. Hence w = 2p /T = 2pf, where f is the frequency of oscillation. Just as with angular velocity, w is measured in radians per second. When wT = 2p radians the trignometric functions recycle. The motion is oscillatory with frequency f = w/2p where k/m .
Simple Harmonic Motion
Simple harmonic motion results when an object is subject to a linear restoring force and is called simple harmonic motion, SHO. The general solution is then:
x Acos t B sint with
k/m
What about the unknown constants A and B? They are determined by the initial conditions. For example if at t = 0 the system satisfies x = xo and v = 0, then:
xt 0 A x o and v dx/dt B 0
The solution for this initial condition becomes:
xt x o cos t
Simple Harmonic Motion
Simple harmonic motion results when an object is subject to a linear restoring force and is called simple harmonic motion, SHO. The general solution is then:
x Acos t B sint with
Graphically this solution is given by:
k/m
Does the expression for angular frequency
k/m
make sense? What happens when the spring constant k increases, what about the dependence on m?