Wednesday, March 30, 2016

DIAGRAMMING VECTOR FORCES

Once the basics of vectors (magnitude, direction, computations) have been covered, most courses introduce practical applications.  At this point, the rather routine manipulations become more obscure.  The first step in deciphering vector word problems is to design a visual representation —

Step One:  DRAW A PICTURE

This step cannot be too heavily emphasized.  A picture puts the problem in context, allows labeling of key parts, highlights what is missing, and puts the problem solver in control of how the relationships are depicted.  Having a standard framework from which to operate is most important in controlling the problem.  Here are several routine situations and common patterns for designing descriptive visual aides.


GIVEN TWO OR MORE VECTORS, DETERMINE THE RESULTANT

The most simple situation yields easily to a ‘Head to Tail’ approach and uncomplicated vector computations.  The design simply puts the back end of the second vector at the arrowhead of the first.  

  Head to Tail --->     

Converting multiple vectors to Head to Tail...

--->     ---> 

Word problems often involve opposing forces: airplane encountering crosswinds, crossing a river with a current, 2 people pulling a cart in different directions....what is the resultant vector, speed (force), direction?


EXAMPLE:  Plane traveling N45E at 200 miles per hour; wind blowing N60E at 100 miles per hour.

  RESULTANT VECTOR
   < 1002 + 50√3, 1002 + 50 >
       ≈ < 228.0, 191.4 >
  SPEED
    √((228.0)^2 + (191.4)^2) ≈ 297.7mph      
  DIRECTION
     Arctan (191.4/228) ≈ 40.0˚
     40˚ = N50E

NOTICE that 40˚ is measured counterclockwise (anticlockwise) from the x-axis as in a trigonometric unit circle (since that's what my calculator does)...but the flight direction is a bearing measured from facing North and turning in an Easterly direction, so the correct answer would be in bearing form.


EXAMPLE:  Crossing river in motorboat going straight east at 20 mph; current flowing straight south at 5 mph.  What is the actual speed and direction of the boat?

   RESULTANT VECTOR
     < 20,5 >
   SPEED
     √[20^2 + (-5)^2 ] ≈ 20.6 mph
   DIRECTION
     bearing S76E = 104˚ from true North

EXAMPLE:  One child pulling a wagon on vector < 10,10 > while another pulls on vector < 5, 5√3 >.  What direction will the wagon move?
First, orient the forces with the coordinate    
plane.

Place vectors head to tail.   

Recognize the task-at-hand is vector addition. 

  RESULTANT VECTOR
    < 15, 10+5√3 > ≈ < 15, 18.7 >
  DIRECTION
    Arctan (18.7/15) ≈ 51.3˚ measured from x-axis


 A variation on the basic question might ask how much force is required to PREVENT movement of the wagon.  Instead of the force applied in the pulling direction, the force would be in the OPPOSITE direction.  In the wagon example, a third child would need to pull with the resultant force at 231.3˚.
 

    

This variation introduces....GIVEN OPPOSING FORCES

This "weighty" problem frequently involves a weight hanging from ropes, guy wires holding up a pole against the wind, or people pulling in counterforce.  This drawing requires the tensions exerted by the upward forces to equal the downward force of the object (wind, third person pulling).


EXAMPLE:  A 40 pound weight is hung from ropes at 30˚ and 45˚ angles from the horizon.

  U = right side rope - 45˚ angle
      vector U = 3 cos 45˚ + 3 sin 45˚
  V = left side rope - 30˚ angle
      vector V = 7 cos 30˚ + 7 sin 30˚
  W = hanging object

    U + V = W

NOTICE that the smaller the angles, the greater the tension on the ropes. 
 

   GIVEN A RAMP

In researching this blog post, I ran into more "ramp" problems than any other kind, many of them in Physics examples.  Because of the ubiquitous problem type, there are many diagrams, each with a special twist.  For use as a data gathering structure, the ramp, load, and solution triangle should be controlled by the problem solver.  If the problem offers a drawing with an unfamiliar layout, I tend to redraw it to my own liking.

Ramp problems require more description of the forces applied.  Gravity, of course, tries to pull the object down the ramp, but the force is actually PERPENDICULAR to flat ground and equal to the weight of the object.  



EXAMPLE:  What force is required to prevent a 100 pound block from sliding down a ramp set at 24 degrees from horizontal? 

                       SIN 24 = x/100
                    x = 40.7 foot pounds



My Physics friends use a more complicated version of the ramp diagram.  It flips the triangle and places the force to push or pull the object or prevent it from slipping down the ramp on a line PARALLEL to the ramp, with the angle touching the inclined plane equaling the angle of elevation of the ramp.

  I offer it here only as an alternative which may have some benefit in Physics.


Before publishing this post, I asked the Physics tutors to review and edit the work for accuracy.  Every one of them commented on the indisputable need to draw the picture!!  Before I added diagrams for each example, every Physics expert scribbled a drawing next to the problem, labeled it, and verified my calculations.  I figure if these experts need the visual aides, we should all take a lesson from them.