Question

A cyclist is riding a bicycle at a speed of 22 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. The shortest distance in which the cyclist can stop without being thrown over the front wheel is

Answer 1

Answer:

The shortest distance is  $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

     The distance between the axial is  $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$  behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                $\sum _B = 0$

=>             $mg (26) = ma(40)$

=>             $a = \frac{26}{40} g$

Here  $g = 32.2 ft/s^2$

     So     $a = \frac{26}{40} (32.2)$

             $a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

       $v^2 = u^2 + 2as$

 Where  $u = 32.26 ft /s$

             and $v = 0$ since the bicycle is coming to a stop

        $v^2 = (32.26)^2 - 2(20.93) S$

=>      $S = 24.86 ft$