How do you calculate the net torque in a friction torque problem?

[Bones]

Homework Statement

Calculate the net torque about the axle of the wheel shown in the figure. Assume that a friction torque of 0.39 m·N opposes the motion.

http://www.webassign.net/gianpse3/10-52.gif

Homework Equations

The Attempt at a Solution

(0.2m)(20N)(sin135)=2.83
(0.1m)(35N)(sin 135)=2.47
(0.2m)(30N)(sin90)=6
-2.83m·N-2.47m·N+6m·N=0.7m·N
0.7m·N+.39m·N=1.09m·N
Is this how you deal with friction in a torque problem??

 

[LowlyPion]

Homework Statement

Calculate the net torque about the axle of the wheel shown in the figure. Assume that a friction torque of 0.39 m·N opposes the motion.

The Attempt at a Solution

(0.2m)(20N)(sin135)=2.83
(0.1m)(35N)(sin 135)=2.47
(0.2m)(30N)(sin90)=6
-2.83m·N-2.47m·N+6m·N=0.7m·N
0.7m·N+.39m·N=1.09m·N
Is this how you deal with friction in a torque problem??

Not exactly. Torque is the force times the distance away that it acts through. Since all 3 are acting tangentially (perpendicular to radii) there is no Sinθ to consider.

The torque attributed to friction will oppose whatever the 3 Forces X distances sum to. Pick a direction of rotation as positive for example counter-clockwise and add those that rotate in that direction. Subtract any that are clockwise. Whatever the result then you deal with the frictional torque such that it reduces the magnitude but doesn't change the sign of the result. (If it goes through 0 then it stays at 0.)

 

[Bones]

Thanks, that helps a lot!