How does a single pivot side-pull caliper bicycle brake work? The brake cable only pulls on one of the arms, so what is causing the force on the other arm? I understand that it is a normal force due to the cable housing, but why is this force developed?
I'll elaborate on my questions a bit. Let $A$ be the point where the cable is attached, $B$ be the point where the housing pushes on the other arm, $P$ be the point of pivot, $C$ be the point where the first arm touches the rim and $D$ be the point where the second arm touches the rim. $APC$ is then the arm pulled on by the cable and $BPD$ is the arm potentially pushed by the housing.
Let's say the cable pulls at $A$ with some tension force $T$. A normal force $N$ from the wheel must then push on $C$, in order for the torque with respect to $P$ to be zero. We have $N = (|AP|_x/|CP|_yS) $, if $x$ and $y$ are in the horizontal and vertical direction respectively, and $T$ is vertical and $N$ horizontal.
The force $N$ gives a lateral force on the rim and a torque on the wheel. They could be balanced by a normal force of the same magnitude at the other brakepad. By symmetry, $B$ would then have to be pushed by a force of magnitude $T$.
But couldn't they also be balanced at the hub? Does it have to do with the construction of the wheel, i.e. because it's spoked? If the wheel were in the form of a solid cylinder (which would of course be very impractical!), would the situation be different?
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