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.
019. Vector quantities
Question: `q001. Note that this assignment contains 6 questions.
. If you move 3 miles directly east then 4 miles directly north, how far do end up from your starting point and what angle would a straight line from your starting point to your ending point make relative to the eastward direction?
Your solution:
Confidence rating:
Given Solution:
If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees.
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Question:
`q002. When analyzing the force is acting on an
automobile as it coasts down the five degree incline, we use and x-y coordinate
system with the x axis directed up
the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000
Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline?
** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees.
If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. **
It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component.
STUDENT QUESTION
what is the difference between the magnitude and the length, or are the the same. I know that in seed 17.2 the magnitude of the gravitational force was found by f=m*a, 5kg * 9.8m/s^2 = 49N, why is this done differently, was this magnitude using f= m*a because gravitational forces act on the vertical or y component?
INSTRUCTOR RESPONSE
The magnitude of a number is its absolute
value.
When working in one dimension, as with F = m a in previous exercises, the force
was either positive or negative and this was
sufficient to specify its direction. For example for an object moving
vertically up and down, the gravitational force is either positive or negative,
depending on the direction you chose as positive.
When working with a vector in 2 dimensions, the magnitude of the vector is
obtained using the pythagorean theorem with its
components.
When sketching a vector, whether the vector represents a displacement, a force,
a velocity, etc., its magnitude is associated with its sketched length.
While the direction of a vector in one dimension can be specified by + or -, the
direction of a vector in 2 dimensions is now specified by the angle it makes as
measured counterclockwise from the positive x direction.
Your solution:
Confidence rating:
Given Solution:
The x component of the weight vector is 15,000
The y component of the weight vector is 15,000
Note for future reference that it is the -1300
STUDENT QUESTION
####What are these numbers telling me in
terms of a real life scenario…if they’re “more than”
opposite (-14,900N+-1,300N versus 15,000N) the weight in N does it roll, and
“less than” opposite does it stay
INSTRUCTOR RESPONSE
Good questions.
-14,900N and -1,300N are in mutually perpendicular directions so they wouldn't
be added; the calculation -14,900N+-1,300N is meaningless
These quantities are associated with legs of a triangle, and the 15000 N with
the hypotenuse. If you add the squares of -14,900N and -1,300N you get the
square of 15,000 N.
The -14,900 N is perpendicular to the incline and all forces perpendicular to
the incline are balanced by the normal force.
The -1300 N is parallel to the incline, and is the reason the object will tend
to accelerate down the incline.
STUDENT QUESTION
My main problem with the visualizations is where to start with the graph. If you say “draw a line from the origin which will be your hypotenuse and connect a line from the end of this line to x=o to form your vertical leg, and from that point to the origin will be your horizontal leg” I got that… but its when the graph is tilted to certain degrees that I get confused, a protractor might help, but it may be more useful to familiarize myself with the nature of vectors themselves and how they are illustrated given certain quantities like the ones above?
INSTRUCTOR RESPONSE
Pictures are there in the Class Notes,
along with diagrams in some of the q_a_ and query documents.
Try this:
Sketch a set of y vs. x coordinate axes in the usual horizontal-vertical
position.
Sketch a ramp which passes through the origin and makes an angle of 30 degrees
above the positive x axis.
Sketch a vector starting at the origin, pointing vertically downward, to
represent the force exerted by gravity on a mass at the origin. The vector will
extend along the negative y axis. Its angle as measured counterclockwise from
the positive x axis will be 270 degrees.
Now imagine that you gradually rotate the x-y axes about the origin, rotating in
the counterclockwise direction until the x axis runs along the incline.
Through how many degrees will you have rotated the axes?
As the axes rotate, the downward-pointing vector stays put. At the end of the
rotation, when the x axis has become parallel to the incline, by how many
degrees with the negative y axis have rotated?
How many degrees with there then be between the negative y axis and the downward
vector?
What then will be the angle of the downward vector, as measured counterclockwise
from the x axis of the rotated coordinate system?
If you submit a copy of these questions with your best responses, you'll
probably gain a better understanding, and in any case I'll be able to see more
specifically what you do and do not understand.
Self-critique (if necessary):
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Question: `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force?
Your solution:
Confidence rating:
Given Solution:
Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees.
STUDENT QUESTION
is this going counterclockwise -53.13 + 360= 307 degrees?
INSTRUCTOR RESPONSE
-53 degrees in one direction is +53
degrees in the opposite direction.
Clockwise and counterclockwise are opposite directions.
So -53 degrees counterclockwise is the same as +53 degrees clockwise.
Now +360 degrees (that is, 360 degrees measured counterclockwise from the
positive x axis) takes us all the way back to the positive x axis.
53 degrees clockwise from that point takes us back to +307 degrees.
-53 degrees counterclockwise from that point does the same.
In other words, there is just one direction, which can be specified as 307 degrees in the counterclockwise direction from the positive x axis, or as -53 degrees counterclockwise from the positive x axis (-53 deg counterclockwise being the same as 53 deg clockwise).
STUDENT QUESTION
I don’t understand why the total force
exerted on the object wouldn’t be just the addition of the two forces. So,
the
magnitude, or the hypotenuse, is the total force exerted on the object in both
directions?
INSTRUCTOR RESPONSE
The net effect of the force is the
resultant of the two components, and unless one of the components is zero the
magnitude of the resultant is less than the sum of the magnitudes of the
components.
This happens for the same reason the hypotenuse of a right triangle is shorter
than the sum of the legs, i.e., the shortest distance between two points is a
straight line.
I'm not going to create a lot of confusion and spell it out in detail here here,
but the connection between distances and forces is pretty direct: velocity is
rate of change of position, acceleration is rate of change of velocity, and net
force = mass * acceleration. Even if the force isn't accelerating anything, it's
the same as it would be if it was.
qed, if we just fill in a few more details</h3>
IRRELEVANT BUT INTERESTING
The following appeared in at least one student's submission. I know the student didn't go to the trouble to insert the apostrophes (and I'm not sure I got the spelling of the plural right just now):
"If I exert a force of 200 Newton’s at an
angle of 30 degrees, and you exert a force of 300 Newton’s at an
angle of 150 degrees, then how great will be our total force and what will be
its direction?"
my response: it's irrelevant but interesting that a word processor
apparently decided to take it upon itself to replace "Newtons" with "Newton's",
its grammar checker thereby making the sentence completely ungrammatical
Self-critique (if necessary):
Self-critique rating:
Question:
`q004. If I exert a force of 200
Your solution:
Confidence rating:
Given Solution:
My force has an x component of 200
Your force has an
x component of 300
In the x direction and we therefore have forces of 173
Newtons and -260 Newtons, which add
up to a total x force of -87 Newtons.
In the y direction we have forces of 100
The total force therefore has x component -87
The magnitude of
the force is `sqrt( (-87
The angle at which the
force is directed, as measured counterclockwise from the positive x axis,
is arctan (250
STUDENT QUESTION
why here are we subtracting from 180 instead of 360?
INSTRUCTOR RESPONSE
We are adding 180 degrees to
the -70.87 degrees. We can look at this in two ways:
1. That's the rule. When the x component of a vector is negative, we add 180
degrees to the arcTangent.
2. The reason for the rule is that the arcTangent can't distinguish between
a second-quadrant vector and a fourth-quadrant vector (we are
taking the arcTan of a negative either way), or between a first-quadrant and a
third-quadrant vector.
Consider a second-quadrant vector whose x component is -5 and y component is +4, and the fourth-quadrant vector whose x component is +5 and whose y component is -4. It should be clear that these vectors are equal and opposite, so that they are directed at 180 degrees from one another.
Now calculate the angles, using the arctangent.
One way you will calculate arcTan(5 / (-4) ) = arcTan(-.8), the other way you will calculate arcTan(-4 / 5) = arcTan(-.8). Both ways you get arcTan(-.8), which gives you about -39 degrees.
If the vector is in the fourth quadrant, as is the case if the x component is +5, this is fine. -39 deg is the same as 360 deg - 39 deg = 321 deg, a perfectly good fourth-quadrant angle.
However if the vector is in the second quadrant, as is the case if the x component is -5, the angle is 180 degrees from the fourth-quadrant vector. 180 deg - 39 deg = 141 deg.
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Question: `q005. Two objects, the first with a momentum of 120 kg meters/second directed at angle 60 degrees and the second with a momentum of 80 kg meters/second directed at an angle of 330 degrees, both angles measured counterclockwise from the positive x axis, collide. What is the total momentum of the two objects before the collision?
Your solution:
Confidence rating:
Given Solution:
The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second.
The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second.
The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second.
The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately.
The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately.
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
Question: `q006. An 80 kg person sits on a 15-degree incline. On a coordinate system whose x axis is directed parallel to the incline, rising from left to right, the vector representing the person's weight makes an angle of 255 degrees as measured counterclockwise from the positive x axis.
What are the x and y components of the person's weight?
To keep the person from sliding down the incline, friction has to exert a force up the incline. How much frictional force is required?
Your solution:
Confidence rating:
Self-critique (if necessary):
Self-critique rating: