Open QA 19

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course PHY 121

4.14.11 at 4:50pm

019. Vector quantities

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Question: `q001. Note that this assignment contains 5 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?

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Your solution:

3^2 + 4^2 = c^2

9 + 16 = c^2

c = 5

angle = tan^-1(4/3)

= 51.13

confidence rating #$&*:

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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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Self-critique (if necessary):

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Self-critique rating:

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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 Newton weight of the car is represented by a vector acting straight downward. This

puts the weight vector at an angle of 265 degrees as measured counterclockwise from the

positive x axis. What are the x and y components of this weight vector?

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 represends 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.

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Your solution:

X component = 15,000cos(265) = -1307.34

y component = 15,000 sin (265) = -14942.92

confidence rating #$&*:

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Given Solution:

The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons

approximately.

The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900

Newtons.

Note for future reference that it is the -1300 Newtons acting in the x direction, which is

parallel to the incline, then tends to make the vehicle accelerate down the incline. The

-14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or

compress the incline, which if the incline is sufficiently strong causes the incline to

exert a force back in the opposite direction. This force supports the automobile on the

incline.

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.

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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?

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Your solution:

400+300 = 700 N in the direction of the negative x axis.

confidence rating #$&*:

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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

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

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Self-critique (if necessary):

I had the same thought process as a student above. I thought you could simply add the two forces together to get the total force on the object. Should have known it wouldn't be that simple!

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Self-critique rating:

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Question: `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert

a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force

and what will be its direction?

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Your solution:

200cos(30)=173.21N

200sin(30)=100 N

300cos(150)= -259.81N

300sin(150)= 150 N

confidence rating #$&*:

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Given Solution:

My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons

approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This

means that the action of my force is completely equivalent to the action of two forces,

one of 173 Newtons in the x direction and one of 100 Newtons in the y direction.

Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y

component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action

of your force is completely equivalent the action of two forces, one of -260 Newtons in the

x direction and one of 150 Newtons in the y direction.

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 Newtons and

150 Newtons, which add up to a total y force of 250 Newtons.

The total force therefore has x component -87 Newtons and y component 250 Newtons. We

easily find the magnitude and direction of this force using the Pythagorean Theorem and the

arctan.

The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons,

approximately.

The angle at which the force is directed, as measured counterclockwise from the positive x

axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg.

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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Self-critique (if necessary):

I wasn't sure where to go from where I left off. Why is it ok to add the x components here but not ok to add the forces acting on the object above?

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Self-critique rating:

@& The x components are in the same direfction (the x direction). Actually they can be in opposite directions (one can be positive and one negative) but the + and - signs take care of that.

In the preceding the forces were not acting along the same line.*@

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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?

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Your solution:

momentum = mass * velocity

confidence rating #$&*:

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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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Self-critique (if necessary):

As with the majority of all other aspects of this class, I am failing to put something together here. As always, once I read your solution, it usually makes sense. I thought that I needed a mass to get started but I guess that kg m/s is what I was looking for. Had I picked up on that, maybe I could have gotten closer to the actual solution.

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Self-critique rating:

@& The thing to pick up on here is that you have to quantities with the same units (in this case kg m / sec), and both quantities have directions.

Thus the two quantities are vector quantities (quantities that have magnitude and directions), and since they have the same units they can be added as vectors.

When you add vectors, you break them into x and y components. Then (per my preceding note) you can add the x components, and you can add the y components.*@

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#