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Phy 201
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.
** CQ_1_19.2_labelMessages **
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: ->->->->->->->->->->->-> : I assume the questions are below.
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What are the x and y components of the equilibrant of the force?
answer/question/discussion: ->->->->->->->->->->->-> : It looks like I know the y but not the x or hypotenuse. If y is -10N, that means the hypotenuse = -10N/sin250 = -10.64 N. And, x = -10.64*
(cos 250) = -3.6. If I take sq root -3.6^2 + -10^2, I confirm the -10.64 magnitude (hypotenuse). So, the equilibriant x and y components would be 10N and 3.6N.
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Once the axes are rotated, the vector no longer lies on the y axis and its y coordinate is no longer -10 N.
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The magnitude of the vector doesn't change when the axes are rotated. However its angle with the positive x axis does change.
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&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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This should be easy to revise. If not don't get hung up on the problem; rather resubmit with questions.
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