********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 3 miles going east is the x-component and the 4 miles going north is the y-component, so the distance from the starting point to the ending point is: c = sqrt(3^2 + 4^2) = sqrt(25) = 5. The angle between the straight line, and the eastward direction or also called the x-axis is: tan-1 (4/3) = 53.13 degrees. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*ent: 3 ********************************************* 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 imagine 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 they 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ok, I sketched my vector, which is only 5 degrees left of the y-axis. I am not sure how to get the x and y components from just the angle of the vector?????? confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline. STUDENT QUESTION ####What are these numbers telling me in terms of a real life scenario…if they’re “more than” opposite (-14,900N+-1,300N versus 15,000N) the weight in N does it roll, and “less than” opposite does it stay INSTRUCTOR RESPONSE Good questions. -14,900N and -1,300N are in mutually perpendicular directions so they wouldn't be added; the calculation -14,900N+-1,300N is meaningless These quantities are associated with legs of a triangle, and the 15000 N with the hypotenuse. If you add the squares of -14,900N and -1,300N you get the square of 15,000 N. The -14,900 N is perpendicular to the incline and all forces perpendicular to the incline are balanced by the normal force. The -1300 N is parallel to the incline, and is the reason the object will tend to accelerate down the incline. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, I understand it now. It’s making better since. ------------------------------------------------ Self-critique rating #$&*ent: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The total force exerted on the object is 300 N + 400 N = 700 N. Obviously, the direction of the force is going to towards the y-axis, because it exerting more force on the object than the other force. The angle of the direction is: tan-1(-400/300) = -53.13 degrees. Since, the angle would be -53.13 degrees in the clockwise direction, you would subtract 360 degrees from the 53.13 degrees to get the counter-clockwise angle: 360 – 53.13 = 306.87 degrees. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 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???
.............................................
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 understand completely that after you get the x and y components on each force, you then add the two together and then get the direction. I was confused about that, but I am not know. ------------------------------------------------ Self-critique rating #$&*ent: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First I need to find the x and y components of both of the vectors: For the 60 degree vector: x-component = 120 cos(60) = 60 kg m/s y-component = 120 sin(60) = 103.9 kg m/s For the 330 degree vector: x-component = 80 cos(330) = 69.3 kg m/s y-component = 80 sin(330) = -40 kg m/s Now, I add the x components and the y components together: x-component: 60 kg m/s + 69.3 kg m/s = 129.3 kg m/s y-component: 103.9 kg m/s + -40 kg m/s = 63.9 kg m/s Now, I can get the total momentum of the object: Total momentum = sqrt(63.9^2 + 129.3^2) = 144.23 kg m/s confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*ent: 3 "