Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Seed quest 13.1
This isn't the 13.1 'seed' question as it appears on the homepage. It was at one time listed for this assignment, but that was changed weeks ago, and this problem was moved to a later assignment. Perhaps you accessed this problem at an earlier date (before 7/18), which would explain how you got it.
You need the basics of vectors, which are given in Asst 16, to do this problem. However see my notes, many of which you will understand, to your future benefit. Don't worry about the parts you don't understand and don't try to answer questions you aren't yet prepared to answer, but file this away for future reference.
A 5 kg cart rests on an incline which makes an angle of 30 degrees with the horizontal.
Sketch this situation with the cart on the incline and the x axis parallel to the incline. The gravitational force on the cart acts vertically downward, and therefore has nonzero components parallel and perpendicular to the incline. Sketch the x and y components of the force, as estimate the magnitude of each component.
answer/question/discussion:
I’m glad this is just a seed question and I am not expected to already know this very well. Since the slope is small the force parallel to the incline (x axis) is approx. equal to the slope * the weight. .03*5kg*9.8m/s^2 = 1.47. Friction also has to be accounted for, so you multiply 5Kg*9.8m/s^2* the given percentage of the weight that is the friction component. Subtract the second product from the first to get the parallel force. The perpendicular force (y) is balanced (meaning 0) because the incline exerts a force perpendicular to itself upward to balance the gravitation force downward.
A 30 degree incline does not have a slope of .03, and the slope cannot be considered small.
The slope of an incline is tan(theta), where theta is the angle with horizontal. You aren't expected to know this at this point, but it's worth remembering.
You are advised to sketch the following as you read it.
The components are calculated as indicated in the instructions, by orienting the x-y coordinate system so that the x axis is parallel to the incline. To do this the 'standard' x-y system with x axis horizontal and y axis vertical is rotated 30 degrees. If the incline is considered to slope downward and to the right, the x axis will be rotated 30 degrees in the clockwise directions, and will end up directed 30 degrees below horizontal. At the same time the y axis will rotate 30 degrees clockwise from vertical. The positive y axis ends up tilted 30 degrees clockwise from vertical, and the upward vertical direction will now lie 120 degrees in the counterclockwise direction from the positive x axis. The negative y axis ends up at 300 degrees as measured counterclockwise from the positive x axis, or 60 degrees as measured clockwise from the positive x axis.
No matter how the coordinate system is rotated, the weight of the cart is a force which acts in the downward vertical direction. So with the new orientation of the axes, the weight vector is at 300 degrees, as measured counterclockwise from the positive x axis.
Your sketch should show the rotated x-y axes, with the downward vertical weight vector, with projection lines from the end of the vector to the positive x axis, and to the negative y axis. The projection line to an axis should be perpendicular to that axis.
Using the definitions of the sine and cosine, find are the components of the cart's weight parallel and perpendicular to the incline.
Should read 'Using the definitions of the sine and cosine, find the components of the cart's weight parallel and perpendicular to the incline.' Editing error when I changed 'what are' to 'find'.
The x and y components of a vector v are v_x = v cos(theta) and v_y = v sin(theta), where theta is the angle measured in the counterclockwise direction from the positive x axis.
What is the magnitude of the weight vector?
What is its angle as measured counterclockwise from the positive x axis?
What therefore are the x and y components of the weight vector?
Compared to the length of your sketched weight vector, do these components look about right?
answer/question/discussion:
I don’t have a clue. First there is a typo in the sentence above, so I am not sure what you are asking. Second I don’t remember Trig. I looked up sine and cosine online and…
The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. Opposite / hypotenuse
The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. Adjacent / hypotenuse
Sine is first, rise is first. Sine takes an angle and tells the rise. Cosine is second, run is second. Cosine takes an angle and tells the run.
So, taking a wild guess. Cosine will tell me how long the incline is (the parallel component) and Sine will tell me the vertical distance that the cart will travel ( the perpendicular component).
How much elastic or compressive force must the incline exert to support the cart, and what is the direction of this force?
answer/question/discussion:
The incline has to exert a force equal and opposite to the vertical gravitational force of the cart. The negative of the mass of the cart * 9.8m/s^2. The incline exerts an upward force.
If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion:
Since the slope is small the force parallel to the incline is approx. equal to the slope * the weight. .03*5kg*9.8m/s^2 = 1.47N
Accel = Fnet/m = 1.47N/5kg = .294 = .3m/s^2
Wouldn’t it be easier to just find the acceleration by multiplying the slope *gravity?
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1 hour
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Be sure you see my note at the beginning of the assignment. This exercise was moved to a later assignment from Asst 13 on or before 7/18/08.