Calculate the net torque about the axle of the wheel

[confusedbyphysics]

This problem is due on my webassign homework in 3 hours and I can't get it right!

Calculate the net torque about the axle of the wheel shown in Fig. 8-39. Assume that a friction torque of 0.43 mN opposes the motion and that F = 26.

The picture is here: http://www.webassign.net/giancoli/8-39alt.gif

What I did:

The 35 N force will move the wheel clockwise so the torque will be -

28 N force will move it counterclockwise so torque will be +

26 N force will move it clockwise so force will be -

They forces are all perpendicular to the radius so its just T = r X F

35 N force T = 35 N X .12 m = -4.2 (moves cw)

28 N force T = 28 X .24 m = 6.72 (moves ccw)

26 N force T = 26 X .24 m = -6.24 (moves cw)

So then including the friction torque which opposes the motion of .43 I put

6.72 - 4.2 - 6.24 - .43 = -4.15 is the net torque and because its - it movies in the clockwise direction

Web assign says -4.15 is WRONG! What am I doing wrong, please help!

 

[lightgrav]

You haven't said yet which direction (cw , ccw) the disk is rotating...
if it was already rotating ccw, the friction torque would be cw.
have you tried that?

 

[whozum]

It says 0.43 mN, which isn't the same as 0.43N

 

[confusedbyphysics]

Ohhhh...so if the motion is going clockwise than the friction will be counter clockwise so it would be +.43 instead of -.43?

6.72 - 4.2 - 6.24 + .43 = -3.29

is that right?? ( I only have one guess left on my webassign and I don't want to screw it up, lol)

 

[lightgrav]

Well, I don't know the whole scenario of the problem, but ...
the 3 forces in the diagram MIGHT be slowing the spin of the disk.
There should be some indication of the spin, in the problem text.

I expect your .43 mN means meter x Newton (not milliNewton)
since it is a torque - unless it is 43 milliN applied at the .12m hub!

 

[Cyrus]

Seems right.

 

[confusedbyphysics]

Ok I just guessed and -3.29 is right. That was stupid of me not to catch that the friction is positive. Thanks for the help!