Phy 231
Your 'cq_1_17.2' 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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A 5 kg cart rests on an incline which makes an angle of 30 degrees with the horizontal.
Sketch this situation with the incline rising as you move to the right and the cart on the incline. Include an x-y coordinate system with the origin centered on the cart, with the x axis directed up and to the right in the direction 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.
What angle does the gravitational force make with the positive x axis, as measured counterclockwise from the positive x axis? Which is greater in magnitude, the x or the y component of the gravitational force?
answer/question/discussion:
the gravitational force acting perpendicular to the x axis, makes an angle of 270 degrees as measured counter clockwise from positive x axis
the magnitude of the gravitational force of the x component is greater than the gravitional force of the y component; this was determined by using the equations Fx = F * sin(theta) and Fy = F * cos(theta)
Using the definitions of the sine and cosine, find the components of the cart's weight parallel and perpendicular to the incline.
answer/question/discussion:
parallel to the incline:
theta = 30 deg
Fx = F * sin(theta)
Fx = 5 kg * 9.8 m/s^2 * sin(30)
Fx = 49 kg m/s^2 * 0.5 = 24.5 N
Fy = F * cos(theta)
Fy = 5 kg * 9.8 m/s^2 * cos(30)
Fy = 49 kg m/s^2 * 0.87 = 42.63 N
F = sqrt( Fx^2 + Fy^2 )
F = sqrt( ( 24.5 N )^2 + ( 42.63 N )^2 )
F = sqrt( 600.25 N^2 + 1817.32 N^2 )
F = sqrt( 2417.57 ) = 49.17 N
perpendicular to the incline:
theta = 90 deg
Fx = F * sin(theta)
Fx = 5 kg * 9.8 m/s^2 * sin(90)
Fx = 49 kg m/s^2 * 1 = 49 N
Fy = F * cos(theta)
Fy = 5 kg * 9.8 m/s^2 * cos(90)
Fy = 49 kg m/s^2 * 0 = 0 N
F = sqrt( Fx^2 + Fy^2 )
F = sqrt( ( 49 N )^2 + ( 0 N )^2 )
F = sqrt( 2401 N^2 + 0 N^2 )
F = 49 N
This seems to indicate that the vertical and horizontal components of gravity are nearly equal...or *are* equal and my totals contain rounding errors.
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 must exert 49 N of force in the upward direction to equal the amount of force (weight) the cart is pushing against the incline with.
If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion:
If no other forces act against the cart, the cart will accelerate at 9.8 m/s^2
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15 min
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The sketch you described has the x and y axes in the horizontal and vertical orientation. Your answers are correct, but they don't follow from the requested sketch, which is consistent with the circular model used in many of my explanations. It's important that you understand how that sketch is constructed and what it tells you.
Your solution did correctly use the equations you get from considering right triangles. It's worthwhile to understand how the right-triangle model and the circular model are related.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#