#$&* course Phy 231 3/11 10pm 019. Vector quantities
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fgrav_x=sin(5deg)*(15000N)=1307.34N Fgrav_y=cos(5deg)*(15000N)=14942.9N 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. 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): It is so much more intuitive for me to envision this with a triangle sort of wedged under the car: the vertical line from the car to ground level is the hypotenuse. The bottom of the triangle extends 5deg above the ground level to the right, parallel to the incline. Then the height of the triangle extends perpendicular to that, back up to the car. I can see logically how an x and y component act in tandem for the resulting hypotenuse, the weight. But this might cause me trouble because I didn't end up with the right signs. Why? How can that be? Is it possible to look at this like a triangle, like I do, but then apply the signs just based on the logic that it is pointing in the negative x direction and the negative y direction? ------------------------------------------------ Self-critique 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F1_x=cos(30deg)*(200N)=173.2N F1_y=sin(30deg)*(200N)=100N F2_x=cos(150deg)*(300N)=-259.8N F2_y=sin(150deg)*(300N)=150N Ftotal_x=173.2N-259.8N=-86.6N Ftotal_y=100N+150N=250N Ftotal=sqrt[(-86.6N)^2+(250N)^2] Ftotal=sqrt(7500N+62500N) Ftotal=sqrt(70000N) Ftotal=264.6N Theta=arctan(250/-86.6) Theta=-70deg 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I knew a -70deg angle did not make sense, but I forgot the rule and couldn't figure out where I went wrong. This is the sort of thing that makes me want to draw triangles, calculate angles on the basis of triangle-based trigonometry, and then assign signs based on logic. But I'm sure it will get more complex that this and there would be times there might not be a particularly obvious visual logic to figure out signs. ------------------------------------------------ Self-critique 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Weight_x=cos(255deg)*(80kg*9.8m/s^2)=-202.9N Weight_y=sin(255deg)*(80kg*9.8m/s^2)=-757.3N Friction must exert enough force to counteract the x-component (parallel to incline) of the person's weight. So frictional force must equal at least 202.9N up the incline. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!