Which disks are in equilibrium?

[taliaroma]

Figure 12-16 shows four overhead views of rotating uniform disks that are sliding across a frictionless floor. Three forces, of magnitude F, 2F, or 3F, act on each disk, either at the rim, at the center, or halfway between rim and center. The force vectors rotate along with the disks, and, in the “snapshots” of Fig. 12-16, point left or right. Which disks are in equilibrium?



2. \tau_{}net = 0
\tau = r(perpendicular)F



3. So I'm pretty sure that I understand why b and d are not in equilibrium, and I'm pretty sure that I understand why a is in equilibrium. I don't know how to figure out why c is in equilibrium, though. I can figure out that net force is 0, but I don't know why net torque is 0. Using the 1st and 2nd equations listed, I got -

\tau_{}net = 0(2F) + R(F) + R(F) = 2RF

where R is the radius. For c to be in equilibrium, the result of this equation has to be 0. How should I do this?

 

[tiny-tim]

welcome to pf!

hi taliaroma! welcome to pf!

(have a tau: τ )

… So I'm pretty sure that I understand why b and d are not in equilibrium, and I'm pretty sure that I understand why a is in equilibrium. I don't know how to figure out why c is in equilibrium, though. I can figure out that net force is 0, but I don't know why net torque is 0. Using the 1st and 2nd equations listed, I got -

\tau_{}net = 0(2F) + R(F) + R(F) = 2RF

where R is the radius. For c to be in equilibrium, the result of this equation has to be 0. How should I do this?[/b]

Your second equation should be τ = r x F …

if you take moments about the centre of the disc, then r is in opposite directions for the two forces, so the two r x Fs have opposite signs

however, I do it by just looking at it and saying oh that's clockwise, that's anti-clockwise!

(btw, note that you can take moments about any point, and it still comes out zero … so you can take a point which has all the forces on the same side!)

(and of course you're right about a b and d )

 

[taliaroma]

Thank you!