# 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 &= 16\text{ in-lb} / 8\text{ in} \\ f &= 2\text{ lb downward} \\ N &= \dfrac{f}{\mu} \\ N &= \dfrac{2}{0.40} \\ N &= 5\text{ 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: $$P = 3.43\text{ lb}$$

• So, show your working so far. – Solar Mike Jan 14 at 16:04