Download Lesson 2 – Vectors, more motion problems, using computers

Lesson 2 – Vectors, more motion problems, using computers
Vectors
An engineer has to deal with physical quantities. Unlike plain numbers in math, physical
quantities usually have dimensions (mass, length and time). In addition, physical quantities
can have both direction and magnitude.
Scalars
Some physical quantities only have magnitude. Examples of such quantities would be
things like mass and time. Such quantities are called Scalars. Adding scalar quantities is
similar to adding numbers in math. The main difference is that scalar quantities have
dimensions and only scalars with the same dimensions can be added. For example, it does
not make sense to add 1 kilogram to 25 seconds. Furthermore, adding scalars also implies
that you convert the scalar quantities so that they are represented in the same units.
For example, suppose you want to add a half hour to 17 minutes:
1
1
60 min
hour 17 min= hour ×
17 min=30 min17 min=47 min
2
2
hour
So, as long as you change the scalars to have the same units, you can add (or subract)
them just like you add (or subtract) numbers.
Some physical quantities have direction in addition to magnitude
Vectors are physical quantities that have direction as well as magnitude (size). Common
examples would be things like forces and velocities. The gravitational force on an object
points towards the center of the earth. The gravitational force has a magnitude equal to the
object's mass multiplied by the acceleration due to gravity. That force has a definite
direction. This direction must be taken into account if you are trying to balance that force
(and keep that object from falling).
Speed is the magnitude of the velocity. So, speed is a scalar. If you look at the
speedometer in a car, it registers the speed. It is possible to drive around the hairpin curve
on the Pali highway with a constant speed. It is not possible to do so with a constant velocity
(or you will go off the road).
Adding vectors
Because vectors have direction as well as magnitude, they must be added in a different way
from how we add scalars and numbers.
Go through the “Animated intro to adding vectors” link to see a step by step explanation
of adding vectors. As you will see in that PDF file, the standard way of adding vectors is to
1
break the vectors into rectangular components. Adding the rectangular components of a
vector is like adding scalar quantities. So, this simplifies the process of adding vectors.
To add vector A to vector B to obtain a resultant vector R:
1. Break vector A and vector B into their x and y components.
2. Add the x components like scalars. That is R x = A x  B x and R y =A y B y
3. Combine R x and R y to obtain the resultant vector R
Although there are some important details left out, the above three steps summarize how
to add vectors.
Here are some sample problems. To check your answers, you can use the “Tool for
adding vectors” link.
Vector Addition Problems
When solving these problems, sketch the vectors first. Check to see if your results match
your drawing. Do not use the “Tool for adding vectors” link until you have solved the
problem by hand using a calculator.
Vector A is 100 N at 30º above the positive x axis. Vector B is 50 N at 20º above the negative x
axis. Find R , where R = A  B
Vector A is 100 N at 30º to the right of the negative y axis. Vector B is 200 N in the positive y
direction. Find R , where R = A  B
Make up your own problems, solve them, and check with the “Tool for adding vectors” link.
Vector Addition using spreadsheets
In the computer lab, we can set up vector addition using spreadsheets. Spreadsheet
programs are valuable programs for an engineer to use. They can be used for many
engineering calculations. Even more important, spreadsheet programs provide two ways of
helping to visualize a solution to a problem involving calculations.
We can set up a spreadsheet to add vectors. The main thing to keep in mind is that the
spreadsheet treats angles as radians not degrees. So, we have to convert our angles in degrees
to radians. In addition, we need to think of our angles as starting at zero for the positive x
direction, and increasing counterclockwise. This means that the positive y direction is 90
degrees, the negative x direction is 180 degrees and the negative y direction is 270 degrees.
Then, we have to convert those degrees to radians.
The conversion of degrees to radians is done using this formula:
2
degrees×
=radians
180
Particle Motion Problems
Let's return to the previous topic of solving particle motion problems. Consider the following
problem:
A particle undergoes rectilinear motion with a constant acceleration of −10m / s 2 .
The particle has an initial velocity of 8.45 m/ s . At what time(s) will the particle have a
position of 2.65 m ? What will the velocity be at that time or times?
You can solve this problem using the quadratic equation, but let's solve it using the
spreadsheet instead.
Start by coming up with the appropriate equation. The equation we want to start with is:
1 2
x =x o v o t  a t
2
1
2
2.65=08.45 t −10 t Let's bring everything to one side.
2
5 t 2−8.45 t2.65=0
The key manipulation is to move everything to one side of the equation, so that the other
side is zero. This means we have:
f t=5 t 2−8.45 t2.65=0
So solutions are when f t=0
Next, we set up the spreadsheet so that we can evaluate
f (t ) .
Note in the formula bar, how we have used the formula:
= 5*(A2)^2 – 8.65*(A2) + 2.65
Now, all we have to do is add more rows with the time value incremented, to see how
f (t ) changes.
3
The formulas to do this are shown in the figure above. Filling these formulas down
several rows will result in the following values.
Note that f(t) crosses zero between 0.2 and 0.4 seconds, and also betwee 1.2 and 1.4
seconds. The approximate solution lies somewhere in those regions. This can be seen if you
make a chart of the data. The arrows show the location of the approximate solutions.
4
You can get better approximations by decreasing the time increment value. Here are the
data with a time increment of 0.1 sec.
Now, you can see the answer is between 1.3 and 1.4. We can go to a smaller increment ,
like 0.01 sec.
This time, we skip over some times, so that this table does not go over too many rows.
5
Now the approximate answers can be seen to be t = 0.40 sec, 1.33 sec. You could go to even
smaller time increments to get greater accuracy if you wanted.
See if you can figure out how to get approximate answers for the velocity at these times.
(Did you get v =4.45 m/s and v =−4.85 m/s ?)
Suppose you change the problem a bit. The particle still starts off with the same initial
velocity of 8.45 m/ s and still undergoes a constant acceleration of −10 m/ s 2 . This
time, find the velocities when the position is 2.47 m .
Make sure you start with the right formula. (Did you get v =±4.69 m/s ?) In this
case, is solving with a spreadsheet a good idea?
6