Question

Review: Figure P14.75 shows the essential parts of a hydraulic brake system. The area of the piston in the master cylinder is $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{2}}$and that of the piston in the brake cylinder is $\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. The coefficient of friction between shoe and wheel drum is $\mathbf{0}\mathbf{.}\mathbf{50}$. If the wheel has a radius of $\mathbf{34}\mathbf{}\mathit{c}\mathit{m}$, determine the frictional torque about the axle when a force of $\mathbf{44}\mathbf{}\mathit{N}$is exerted on the brake pedal.

Short answer

The frictional torque about the axle when a force of $44N$ is exerted on the brake pedal is $\tau =27N.m$.

Step-by-step solution

A concept:

Pascal’s law states that when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.

Pascal’s principle,

$\frac{{\mathbf{F}}_{\mathbf{1}}}{{\mathbf{A}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{2}}}{{\mathbf{A}}_{\mathbf{2}}}$

Determine the frictional torque about the axle:

Pascal’s principle, $\frac{{\mathbf{F}}_{\mathbf{1}}}{{\mathbf{A}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{2}}}{{\mathbf{A}}_{\mathbf{2}}}$

$$\begin{gathered} \frac{F_{pedal}}{A_{mastercylinder}}=\frac{F_{brake}}{A_{brakecylinder}} \\ {F}_{brake}=\left(\frac{A_{brakecylinder}}{A_{mastercylinder}}\right)F_{pedal} \\ =\left(\frac{6.4c{m}^2}{1.8c{m}^2}\right)\left(44N\right) \\ =156N \end{gathered}$$

This is the normal force exerted on the brake shoe. The frictional force is

$$\begin{gathered} f={\mu }_{k}n \\ =0.50(156N) \\ =78N \end{gathered}$$

And the torque is

$$\begin{gathered} \tau =f·r_{drum} \\ =(78N)(0.38m) \\ =27N.m \end{gathered}$$

Hence, the frictional torque about the axle when a force of $44N$ is exerted on the brake pedal is $27N.m$.