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PHY 121
Your 'cq_1_19.1' 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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The equilibrant of a force is the force which is equal and opposite to that force. If two forces are equal and opposite, their x and y components are also equal, but the x and y components of the force are opposite in sign to those of the equilibrant.
The x and y components of a force are 2 Newtons and 3 Newtons repectively.
What are the magnitude of this force and what angle does it make as measured counterclockwise from the positive x axis?
Mag^2 = 2N^2 + 3N^2
Mag = 3.6N
Theta = arctan(3N/2N)
=56.3d
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What are the components of the equilibrant force?
Equl_x = -(2N) = -2N
Equl_y = -(3N) = -3N
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What angle does the equilibrant force make as measured counterclockwise from the positive x axis?
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The angle of the equilibrant force is arcTan(-3 N / (-2 N) ), plus 180 degrees since the x component is negative.
That will put it equal and opposite the original force, which you correctly calculated as about 56 degrees.
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&#Good responses. See my notes and let me know if you have questions. &#
PHY 121
Your 'cq_1_19.2' 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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Sketch a vector representing a 10 Newton force which acts vertically downward.
Position an x-y coordinate plane so that the initial point of your vector is at the origin, and the angle of the vector as measured counterclockwise from the positive x axis is 250 degrees. This will require that you 'rotate' the x-y coordinate plane from its traditional horizontal-vertical orientation.
answer/question/discussion: ->->->->->->->->->->->-> n:
Since a normal x-y plane will have the -y axis at 270d, this plane will be rotates by 20d to create the angle
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