phy121
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.
So we should have a 10N force at an angle of 250*, correct?
What are the x and y components of the equilibrant of the force?
The XY of the initial force is:
x = 10N * cos(250*) = -3.42N
y = 10N * sin(250*) = -9.39N
Therefore, you can either subtract or add 180* to the angle to get a positive angle, I'll subtract to keep it between 0 and 360*
250* - 180* = 70*
So, you have an equal force of 10N:
x = 10N * cos(70*) = 3.42N
y = 10N * sin(70*) = 9.39N
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10min
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&#This looks very good. Let me know if you have any questions. &#