Two forces, F1=100 N (Newton) and F2=20N, are applied on opposite sides of a door's knob at angles theta_1=30 degrees and theta_2=60 degrees respectively, causing the door to rotate (about the axis passing though its hinges). If the distance from the knob to the rotation axis is 0.8 meter, find the net torque exerted on the door and specify the direction of rotation of the door (in this problem, assume that the force F1 causes a counterclockwise rotation and F2 a clockwise rotation). A.) Net Torque = 26.14 N.m; the door rotates counterclockwise. B.) Net Torque = -25.4 N.m; the door rotates counterclockwise. C.) Net Torque = 123.8 N.m; the door rotates clockwise. D.) Net Torque = 1.784 N.m; the door rotates clockwise.

Principles of Physics: A Calculus-Based Text
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Chapter10: Rotational Motion
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Two forces, F1=100 N (Newton) and F2=20N, are applied on opposite sides of a door's knob at angles theta_1=30 degrees and theta_2=60 degrees respectively, causing the door to rotate (about the axis passing though its hinges). If the distance from the knob to the rotation axis is 0.8 meter, find the net torque exerted on the door and specify the direction of rotation of the door (in this problem, assume that the force F1 causes a counterclockwise rotation and F2 a clockwise rotation).

A.) Net Torque = 26.14 N.m; the door rotates counterclockwise.
B.) Net Torque = -25.4 N.m; the door rotates counterclockwise.
C.) Net Torque = 123.8 N.m; the door rotates clockwise.
D.) Net Torque = 1.784 N.m; the door rotates clockwise.
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