Question 9.2.1: A bicycle rider squeezes the brake lever on the handlebars, ...

Goal Find the normal force acting on the wheel rim at E and the mechanical advantage of this brake system.
Given A coordinate system, along with information about how the brake calipers work, their associated geometry, and the force pulling on wire BGC at G.
Assume The system is planar and in equilibrium (with the brake pads engaged with the wheel). Component weights are negligible, pin A is frictionless, and the connection at G between the cable and the wire is frictionless.
Draw We draw a free-body diagram of the connection at G (Figure 2) to determine the tensile force acting on wire BGC (T_{BGC}). We also need a free-body diagram of the caliper. Fortunately brake symmetry allows us to analyze just one of the calipers. We draw a free-body diagram of caliper BAE (Figure 3) and apply the planar equilibrium equations to determine F_{normal}, the normal force acting on the brake pad at E. F_{normal} acting on the brake pad is equal and opposite to the normal force acting on the wheel rim.
Formulate Equations and Solve First we find T_{BGC}, the wire tension pulling on the caliper at B. The wire can slide freely through the connection at G because there is no friction, therefore the tension acting on the wire to the left of the connection is equal to that acting on the right portion.
based on Figure 2 we write:

\sum{F_{y}\left(\uparrow + \right) } =-T_{BGCy}- T_{BGCy} + 200  N = 0

using the geometry of the wire to calculate the y component, we get:

– T_{BGC}\left\lgroup\frac{5}{6.73} \right\rgroup – T_{BGC}\left\lgroup\frac{5}{6.73} \right\rgroup+ 200  N = 0
T_{BGC}=134.5  N            (1)

We now analyze caliper BAE (Figure 3) to find F_{normal}:

\sum{M_{z @ A} }\left(\curvearrowleft + \right) =- T_{BGCy} (4.5  cm) – T_{BGCx} (2  cm) + F_{normal} (9  cm) = 0

Recognizing that T_{BGCy}=\frac{5}{6.73}T_{BGC}    ,  T_{BGCx}=\frac{4.5}{6.73} T_{BGC} and that T_{BGC} = 134.5  N  (from (1)) we get:

– 134.5  N \left\lgroup\frac{5}{6.73} \right\rgroup (4.5  cm) – 134.5  N\left\lgroup\frac{4.5}{6.73} \right\rgroup (2  cm) F_{normal}(9  cm) = 0 \Rightarrow F_{normal} = 70.0  N

The mechanical advantage of the caliper system is the ratio F_{output}/F_{input}. If we consider that the caliper system is made up of two calipers, the total “output” force is 2 × 70 N = 140 N. The “input” force is 200 N. Therefore the mechanical advantage of the caliper system is:

\frac{140  N}{200  N} =0.70

You may be surprised that this number is less than one. Remember– this is the mechanical advantage of just the brake calipers. The mechanical advantage of the whole brake system would need to include both the caliper and the brake lever.
Check To check our result, we reanalyze the problem by first finding F_{Ay} and F_{Ax} and then F_{normal}. based on Figure 3:

\sum{F_{y}\left(\uparrow + \right) }= T_{BGCy}+ F_{Ay}=0\Rightarrow F_{Ay}= – T_{BGC}\left\lgroup\frac{5}{6.73}\right\rgroup= – 100  N
\sum{M_{z @ E} }\left(\curvearrowleft + \right) =- T_{BGCy}(6.5  cm)- T_{BGCx}(11  cm)- F_{Ay}(2  cm) – F_{Ax}(9  cm) = 0
– 134.5  N \left\lgroup\frac{5}{6.73} \right\rgroup(6.5  cm) – 134.5  N\left\lgroup\frac{4.5}{6.73} \right\rgroup (11  cm) – ( 100  N)(2  cm) -F_{Ax} (9   cm) =0\Rightarrow F_{Ax}=- 160.0  N
\sum{F_{x} \left(\rightarrow + \right) } = T_{BGCx}+ F_{Ax}+ F_{normal}=0
134.5  N \left\lgroup\frac{4.5}{6.73} \right\rgroup +(- 160.0  N)+ F_{normal}=0\Rightarrow F_{normal}= 70.0  N

We get the same result!