If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

SOLUTIONS/COMMENTARY ON QUERY 9

......!!!!!!!!...................................

  ****   Query   problem 7.3.14  b = 4, c = 1, alpha = 120 deg

......!!!!!!!!...................................

                                                  00:36:10

Given Solution: 

a= sqrt(4^2 + 1^2 -2(4)(1)cos120)

a = sqrt(16+1-8cos120)

a = sqrt(17-8cos120)

a = 4.58

b^2 = a^2 + c^2 -2accos`beta

4^2 = 4.58^2 + 1^2 - 2(4.58)(1)cos`beta

9.16cos`beta = 4.58^2 + 1^2 - 4^2

9.16cos`beta = 5.98

beta = cos^-1(5.98/9.16)

beta = 49.2deg

c^2 = a^2 + b^2 -2abcos`gamma

1^2 = 4.58^2 + 4^2-2(4.58)(4)cos`gamma

36.64cos`gamma = 35.98

gamma = cos^-1(35.98/36.64)

gamma = 10.89

.........................................

Self-critique (if necessary):

Self-critique Rating:

 ****   Query   problem 7.3.32  500 ft tower 15 deg slope 2 guy wires 100 ft on either side of base on slope

......!!!!!!!!...................................

                                                  00:45:52

Your solution:  

Confidence Assessment:

Given Solution: 

 **  GOOD STUDENT SOLUTION: 

A triangle containing the tower side and the side of the slope up the hill will have angle 75 deg between tower and hill.   The triangle with the side down the slope has angle 105 deg between tower and hill.

The guy wire will form the hypotenuse c on each side.

If

c1 = length of guy wire on the right hand side of the tower and

c2 = length of guy wire on the left hand side of the tower

then the Law of Cosines gives us

(c1)^2 = a^2 + b^2 - 2ab cos `gamma

       = 500ft^2 + 100ft^2 - 2(500)(100)(cos75 deg)

       = 260,000 - 25,881.90451

(c1)^2 = 234118.0955

c = approx. 483.8575 feet

(c2)^2 = a^2 + b^2 - 2ab cos`gamma

        = 500ft^2 + 100ft^2 -2(500)(100)(cos105 deg)

        = 260,000 - (-25881.90451)

(c2)^2 = 285,881.9045

c =  approx. 534.68 feet **

Self-critique (if necessary):

Self-critique Rating:

 ****   query   problem 7.5.6 coiled spring 10 cm displ period 3 sec moving downward at equil when t = 0.

What is the equation of motion of the object?

......!!!!!!!!...................................

                                                  20:29:55

Your solution:

Given Solution:

 ** Modeling with reference-circle point moving counterclockwise at angular velocity omega on a circle of radius A, and using the sine function, displacement will be

y(t) = A sin (`omega * t + theta0), where theta0 is the initial angular position on the reference circle.

We know that the spring is at equilibrium when t = 0, so theta0 is either 0 or pi.

Since the spring is initially moving downward, we conclude that theta0 is pi.

The amplitude of motion will be 10 cm so A = 10 cm.

The period of motion is 2`pi/`omega = 3 sec, so `omega = 2`pi/(3 sec), so

the equation of motion is y = 10 cm sin( 2 pi / (3 sec) * t + pi).

Alternatively we could model using the cosine function, though this is less intuitive for an object moving up and down:

x(t) = A cos(omega * t + theta0),

in which case since the spring is at equilibrium when t = 0, theta0 is pi/2 (motion at t = 0 therefore being in the negative direction)

and we get

x(t) = 10 cm cos(2 pi / (3 sec) * t + pi/2).

STUDENT COMMENT:   I don’t understand why + pi is necessary on the end

INSTRUCTOR RESPONSE:  One stated condition is that the object is moving downward when t = 0. The equation you gave would have had it moving upward.

In order for the object to be at equilibrium, with a sine model, the angle on the reference circle must be either 0 or pi. If the angle is 0, then since the positive direction is counterclockwise (e.g., from the positive x axis into the first quadrant), the sine function will immediately move into positive values, indicating upward motion.

If the angle is pi, then continued positive motion around the reference circle will take you into the third quadrant, where the sine takes negative values; this indicates downward motion.

......!!!!!!!!...................................

Self-critique (if necessary):

Self-critique Rating:

  ****   query   problem 7.5.12 d=5 cos(`pi /2)t

Describe the motion of the object

Your solution: 

Confidence Assessment:

Given Solution:  

** This equation is of form d = A cos(omega * t) with A = 5 and omega = pi/2, corresponding to motion on a reference circle of radius 5 at angular velocity pi/2 rad/sec.

The period of motion is 2 pi / omega = 2 pi / (pi/2) = 4.  So frequency is 1 / period = 1/4.

The amplitude of motion is 5 units, the coefficient of the cosine function. **

.........................................

Self-critique (if necessary):

Self-critique Rating:

****   query   problem 7.4.14 b = 4, c = 1, alpha = 120 deg

What is the area of the triangle?

Your solution:

Given Solution:

 ½ b c sin alpha

.5(4)(1)sin120 = 1.73

Self-critique (if necessary):

Self-critique Rating:

 ****   query   problem 7.4.28 cone from circle 24 ft diameter, 100 deg sector removed

Find the area of the cone obtained when you fold join the cut edges.

......!!!!!!!!...................................

                                                  00:56:16

Your solution:

Confidence Assessment:

Given Solution:

100deg = 1.745rad

Area of Circle = pi*r^2 = pi*12^2 = 452.389

Area of Sector = 1/2(r^2)(`theta rad) = 1/2(12^2)(1.745) = 125.64

Area of Cone = 452.389 - 125.64 = 326.75

.........................................

Self-critique (if necessary):

Self-critique Rating: