Calculate the net torque?

Question

Calculate the net torque about the axle of the wheel shown in the link below. Assume that a fraction torque of .30 Nm opposes the motion.

http://img293.imageshack.us/my.php?image=circle2me6.jpg

Answer 1

Just multiply all the forces by their respective radii and sum them, making sure you use one polarity for clockwise and the opposite polarity for counterclockwise. For a right-handed coordinate system, if +x is to the right and +y is up, the + torque (+z) axis is "out of the page" or toward you so you are looking along the -z axis. +z rotation is clockwise viewed along the +z axis and thus counterclockwise in the viewing direction.
You have:
F=35 * r=10 = -350 N-cm (CW in view direction)
F=20 * r=20 = -400 N-cm (CW in view direction)
F=30 * r=20 = +600 N-cm (CCW in view direction)
Total due to applied forces = -150 N-cm = -1.5 Nm
Total with opposing torque = -1.5 + 0.3 = -1.2 Nm

Answer 2

Net Torque

Answer 3

You would get 1.78 because the hypotenuse is 0.18 m, and each of the angle the right triangle makes it makes is 45 degrees. Use special right angle trig for 45,45,90. The .18/sqrt(2)=0.127. Then tau=(.127)(14)= 1.78 N•m