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

The shortest distance.

Answer to Problem 16.153RP

The value of shortest distance is, $s=20.5366ft.$

Explanation of Solution

Given information:

Speed=20mph

Distance between the axles=42in.

Center of mass is located at=26in

Explanation:

Here, the condition in which cyclist can stop without being thrown over the front wheel .Here the resultant forces will cause the moment point of front wheel. So we consider vertical reaction of force as zero.

So first calculate the normal reaction at point B.

$\begin{array}{l}\sum F_y=0\\ W-N_B=0\\ N_B=W\\ =mg\end{array}$

Calculation:

And then apply the moment balance at point B

$\begin{array}{l}\sum M_B=0\\ \left(W\times 26\right)-\left(ma\times 40\right)=0\\ \left(mg\times 26\right)-\left(ma\times 40\right)=0\\ a=g\times \frac{26}{40}\\ a=32.2\frac{ft}{s^2}\times \frac{26}{40}\\ =20.93\frac{ft}{s^2}\end{array}$

Calculate the distance by using the kinematics formula

$\begin{array}{l}{v}^2-u^2=-2as\\ u^2-v^2=2as\\ s=\frac{u^2-v^2}{2a}\\ s=\frac{{\left(20mph\frac{1.4666ft/s}{1mph}\right)}^2-0}{2\left(20.93\right)}\\ s=\frac{859.6624\frac{f{t}^2}{s^2}}{41.86\frac{ft}{s^2}}\\ s=\frac{859.6624}{41.86}ft\\ s=20.5366ft\end{array}$