Phy 231
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.
** **
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:
Answer: Just to make sure I’m drawing this right, the y axis is now turned 20 degrees to the negative x direction. The line is then drawn through the y axis.
The force vector remains vertical.
The coordinate system in standard orientation will have the y axis vertical, the x axis horizontal. The force vector will be along the negative vertical axis.
The coordinate system then rotates until the force vector is at the 250 degree position. This will require rotating the coordinate system counterclockwise through an angle of 20 degrees, so that the x axis points in a direction 20 degrees above horizontal, the y axis 20 degrees to the 'left' of vertical. The negative y axis will then be rotated at 20 degrees from the original vector.
• What are the x and y components of the equilibrant of the force?
answer/question/discussion:
Answer: x component of the equilibrant of the force: 0N, y component of the equilibrant of the force: -10N.
The x and y components of a vector of magnitude 10 N, directed at 250 degrees (as measured counterclockwise from the positive x axis) are
x component: 10 N * cos(250 deg) = - 3 N (very approximately) and
y component: 10 N * sin(250 deg) = -9 N (very approximately.
** **
15 minutes
** **
&#Your work looks good. See my notes. Let me know if you have any questions. &#