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

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SUMMARY

The net torque calculation in a friction torque problem involves summing the torques produced by various forces acting on the wheel and then accounting for the opposing friction torque. In this case, the calculated torques from three forces are 2.83 m·N, 2.47 m·N, and 6 m·N, leading to a total of 0.7 m·N before considering the friction torque of 0.39 m·N. The final net torque is 1.09 m·N, which reflects the influence of friction opposing the motion. The correct approach emphasizes that friction torque reduces the net torque's magnitude without altering its sign.

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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??
 
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Bones said:

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.)
 


Thanks, that helps a lot!
 

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