Chapter 8.2, Problem 20PP

To determine

The force exerted by the chain to keep the wheel from turning.

Answer to Problem 20PP

The force exerted by the chain to keep the wheel from turning is $F_{\text{chain}}\text{=}789\text{N}$ .

Explanation of Solution

Given:

The mass of the bicyclist is $m=65\text{kg}$ .

The length of the crank is $r_{\text{1}}=0.170\text{m}$ .

The angle made by the crank with the vertical is $\theta =45°$ .

The radius of the chain wheel is $r_{\text{2}}=9.70\text{cm}$ .

Formula used:

The expression for the torque is,

  $\tau =Fr\sin \theta$

Here, $F$ is the applied force, $r$ is the radius of the wheel, and $\theta$ is the angle made by an object with the vertical.

Calculation:

Consider a crank is connected to a chained wheel.

The crank connected to the wheel is shown in Figure below.

  

The weight of the bicyclist is denoted by $F_{\text{g}}$ .

The force exerted by the chain is denoted by $F_{\text{chain}}$ .

Consider the torque exerted by the weight of the bicyclist is denoted by ${\tau }_{\text{bicyclist}}$ .

Consider the torque exerted by the force of chain is denoted by ${\tau }_{\text{chain}}$ .

Consider the acceleration due to gravity of the Earth is $g=\text{9}\text{.80}{\text{m/s}}^2$ .

The torque exerted by the weight of the bicyclist is,

  $\begin{array}{c}{\tau }_{\text{bicyclist}}=F_{\text{g}}{r}_1\sin {\theta }_{\text{1}}\\ =\left(mg\right)r_1\sin \left(45°\right)\\ =\left(65\text{kg}\times \text{9}\text{.80}{\text{m/s}}^2\right)\times 0.170\text{m}\times \left(\frac{1}{\sqrt{2}}\right)\\ =\left(637\text{kg}\cdot {\text{m/s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\times 0.170\text{m}\times \left(\frac{1}{\sqrt{2}}\right)\end{array}$

  $\begin{array}{c}{\tau }_{\text{bicyclist}}=\left(637\text{N}\right)\times 0.170\text{m}\times \left(\frac{1}{\sqrt{2}}\right)\\ =76.57\text{N}\cdot \text{m}\end{array}$

The torque exerted by the force of chain is,

  $\begin{array}{c}{\tau }_{\text{chain}}=F_{\text{chain}}{r}_2\sin {\theta }_{\text{2}}\\ =F_{\text{chain}}\left(9.70\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sin \left(90°\right)\\ =\left(0.097\text{m}\right)F_{\text{chain}}\end{array}$

For the equilibrium of the chain wheel,

  $\begin{array}{c}{\tau }_{\text{chain}}={\tau }_{\text{bicyclist}}\\ \left(0.097\text{m}\right)F_{\text{chain}}=76.57\text{N}\cdot \text{m}\\ F_{\text{chain}}\text{=}\frac{76.57\text{N}\cdot \text{m}}{0.097\text{m}}\\ F_{\text{chain}}\text{=}789\text{N}\end{array}$

Conclusion:

Therefore, the force exerted by the chain to keep the wheel from turning is $F_{\text{chain}}\text{=}789\text{N}$ .