Chapter 16, Problem 16.153RP

A cyclist is riding a bicycle at a speed of 20 mph on a horizontal road. The distance between the axles is 42 in., and the mass center of the cyclist and the bicycle is located 26 in. behind the front axle and 40 in. above the ground. If the cyclist applies the brakes only on the front wheel, determine the shortest distance in which he can stop without being thrown over the front wheel.

To determine

Find the shortest distance in which the cyclist can stop the cycle without being thrown over the front wheel.

Answer to Problem 16.153RP

The shortest distance is $\underset{\_}{s=20.6\text{ft}}$.

Explanation of Solution

Given information:

The initial speed of the bicycle is $v_0=20\text{mph}$.

The distance between the axles is $42\text{in}\text{.}$

The location of mass center of the cyclist and the bicycle is $26\text{in}\text{.}$ behind the front axle and $40\text{in}\text{.}$ above the ground.

Calculation:

Consider the acceleration due to gravity as $g=32.2\text{ft}/{\text{s}}^{\text{2}}$.

Sketch the system as shown in Figure 1.

The cyclist applies the brake only on the front wheel. Hence, $N_A=0$.

Show the free body diagram of the system as shown in Figure (2).

Refer to Figure 3.

Apply the Equilibrium of moment about B as shown below.

$\begin{array}{l}{\displaystyle \sum M_B}=I_G\alpha +mad\\ mg\times 26=m\overline{a}\times 40\\ \overline{a}=\frac{26}{40}g\end{array}$

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}\overline{a}=\frac{26}{40}\times 32.2\\ =20.93\text{ft}/{\text{s}}^{\text{2}}\end{array}$

The initial velocity $\begin{array}{c}{v}_0=20\text{mph}\times \frac{1.4667\text{ft}/\text{s}}{1\text{mph}}\\ =29.334\text{ft}/\text{s}\end{array}$

Calculate the distance $\left(s\right)$ as shown below.

$v^2-v_0^2=2\overline{a}s$

Substitute $0$ for $v$, $29.334\text{ft}/\text{s}$ for $v_0$, and $-20.93\text{}\text{ft}/{\text{s}}^{\text{2}}$ for $\overline{a}$.

$\begin{array}{l}0-{29.334}^2=2\times \left(-20.93\right)s\\ s=20.56\text{ft}\end{array}$

Therefore, the shortest distance is $\underset{\_}{s=20.6\text{ft}}$.