Multiple torques/forces on a disk, finding net torque

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The discussion revolves around calculating the net torque on a 20-cm-diameter disk rotating on an axle. Initial calculations yielded a net torque of approximately 2.35 Nm, but a peer pointed out that one force was not applied to the disk, leading to a revised calculation of about 0.94 Nm. The importance of correctly applying signs to torque calculations was emphasized, particularly for forces causing counterclockwise rotation. Ultimately, the correct net torque was confirmed to be -0.94 Nm after addressing the sign issues. The focus on proper torque direction and force application is crucial for accurate results in physics problems.
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Homework Statement



The 20-cm-diameter disk in the figure can rotate on an axle through its center.

What is the net torque about the axle?

image: http://session.masteringphysics.com/problemAsset/1000963/5/knight_Figure_13_14.jpg

Homework Equations



τ=Frsinφ

The Attempt at a Solution



τnet= τ1+τ2+τ3+τ4

τ1 = 30N * 0.1 * sin(90)
τ2 = 30N * 0.05 * sin(-45)
τ3 = 20N * 0.05 * sin(-90)
τ4 = 20N * 0.1 * sin(45)
_____________________
τnet =~2.35355 Nm

this was my original solution, which came out incorrect, a friend then told me that τ4 is zero because the force of 20N is not being applied to a point on the disk itself, so then:

τ1 = 30N * 0.1 * sin(90)
τ2 = 30N * 0.05 * sin(-45)
τ3 = 20N * 0.05 * sin(-90)
_____________________
τnet =~0.94 Nm

the issue is that I have only one attempt left at this problem and I want to confirm this before I submit the problem
 
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That looks kike it should be the right answer, but I think it is less confusing and error prone if you keep the signs out of the sin arguments, in other words if a force is tending to cause conterclockwise rotation, make the entire torgue negative,

eg, T2=-30N*.05m*sin(45)
T3=-20N*0.05*sin(90)

Else it looks fine.
 
got it, yea the signs were messing me up, i got 0.94 when its actually -0.94,

thanks
 
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