Force necessary to resist motion due to torque on a drum brake

A drum is subjected to a counterclockwise torque of 16 in-lb. What horizontal force is necessary to resist motion?

Coefficient of friction = 0.40

Weight is neglected.

I’m having a hard time on this one. I’ve tried to separate the torque into a normal and tangential force and do a summation of moments at B and even at A. No luck. Though I doubt that’s how it should be tackled.

Edit (my trials/errors):

I’ve read a pdf on brakes and learned that there are so many types and equations to accommodate each one. The closest I’ve seen to the one above was that of a single block brake; with 3 variations on the position of the line of action (the tangential force), $f$, passing through, above, or below the fulcrum.

![An example of a brake system, with the tangential force below the fulcrum]
(https://i.stack.imgur.com/xtKAl.jpg)

It doesn’t look the same, but the proof in this was the one I used. That being:

$$T_b text{ (Braking torque)} = f text{ (tangential force)} times r text{ (radius of wheel/drum)}$$

And the old

$$f = mu text{ (coefficient of friction)} times N text{ (normal force)}$$

I think my mistake was that I used the 16 in-lb torque as the braking torque ($T_b$) and went on from there.

$$begin{align}
T_b &= fr
therefore f &= dfrac{T_b}{r}
f &= 16text{ in-lb} / 8text{ in}
f &= 2text{ lb downward}
N &= dfrac{f}{mu}
N &= dfrac{2}{0.40}
N &= 5text{ lb leftward}
end{align}$$

And then a summation of moments at point B (counterclockwise positive), giving:

$$begin{align}
sum M_B &= P(9) – N(9) + f(2) = 0
therefore 9P &= 5(9) – 2(2)
P &= 4.5556lb
end{align}$$

I also considered the pinned support reaction at A, which gave off another incorrect answer.

The correct answer is given: $P = 3.43text{ lb}$