Calculating Force with Unequal Hand Distances

[gomerpyle]

Suppose you had a bar that spinned due to a torque load applied, but the pivot was not centered at exactly the middle of the bar. If you arrested the bar with your hands, obviously you are applying a couple to counter the torque to stop it from moving, but your hands are not at equal distances to where the torque load is. If you know the torque load, and the distance from the pivot to where your hands are, how do calculate the force each hand needs applies? If the pivot as at the center, the forces would be equal and the magnitude M/d. I think I'm over-thinking this...

 

[billy_joule]

If the pivot as at the center, the forces would be equal and the magnitude M/d.

You're arms aren't of equal strength so the forces won't necessarily be equal, even if you tried. The fact that one arm will be pushing and one pulling makes it even harder to try to apply equal force, regardless of any difference in lever arm length.
All you can say is that:
T_total = T_LH + T_RH = F_LHr_LH + F_RHr_RH

 

[CWatters]

+1

and that's true even if the pivot is in the middle.

 

[jack action]

Each hand will create its own moment: $M_R = F_R r_R$ and $M_L = F_L r_L$ and the total moment is $M = M_R + M_L$. So you could stop the bar with only one hand if you wish.

The difference between a moment and a couple is that the couple have two equal and opposite forces separated by a distance $d$. So if $F_R = F_L = F$, then $M = Fr_R + Fr_L = F(r_R + r_L) = Fd$.

The resultant moment from a couple is called a torque ($Fd$). The torque has the special property of being independent from the reference point. This means that if you create a couple with your hands that produces a torque equivalent to the bar torque, your hands could be on the same side of the bar and it will still stop the bar.

Ref.: Couple (mechanics)