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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: Is there a question? I don’t understand. I have a graph where my vector is in the 3rd quadrant forming a 70deg reference angle, the magnitude of the vector is 10N.
• What are the x and y components of the equilibrant of the force?
y = 10N*sin(250deg) = -9.4N
x = 10N*cos(250deg) = -3.4N
These are the x and y coordinates of the force. The force is (-9.4 N, -3.4 N).
vector of equilibriant is <3.4N, 9.4N>
this is the correct equilibrant
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5 min
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&#Good responses. See my notes and let me know if you have questions. &#