Chapter 11.3, Problem 64E

To determine

To calculate: The magnitude of torque Ton a crankshaft when downward force of $2000\text{ lb}$ is applied.

Answer to Problem 64E

The magnitude of torque T is $\Vert T\Vert =160\sqrt{3}$ .

Explanation of Solution

Given information:

When brakes are applied in a bicycle with help of downward force of $2000\text{ lb}$ on pedals, the crank which is of $0.16\text{ ft}$ make a $60°$ angle with force, the position vector of the crank is $V=0.16\left(\cos 30°j+\sin 30°k\right)$ and the force is $F=-2000k$ .

Diagram of the above situation is provided below,

  

Formula used:

The magnitude of the torque T on the crank when downward force is applied on pedal is $\Vert V\times F\Vert$ , where $V$ is position vector of crank.

Let $a$ and $b$ be any two arbitrary vectors, then the cross product of the vectors is denoted by $a\times b$ .

Mathematically the magnitude of cross product is expressed as, $\Vert a\times b\Vert =\Vert a\Vert \cdot \Vert b\Vert \cdot \sin \theta$ , where $\Vert a\Vert$ and $\Vert b\Vert$ denote the magnitude of vectors $a$ and $b$ respectively and $\theta$ is the angle between them.

Calculation:

When brakes are applied in a bicycle with help of downward force of $2000\text{ lb}$ on pedals, the crank which is of $0.16\text{ ft}$ make a $60°$ angle with force, the position vector of the crank is $V=0.16\left(\cos 30°j+\sin 30°k\right)$ and the force is $F=-2000k$ .

Diagram of the above situation is provided below,

  

The angle $\theta$ between $V$ and $F$ is.

  $\theta =60°$

Let $a$ and $b$ be any two arbitrary vectors, then the cross product of the vectors is denoted by $a\times b$ .

Mathematically the magnitude of cross product is expressed as, $\Vert a\times b\Vert =\Vert a\Vert \cdot \Vert b\Vert \cdot \sin \theta$ , where $\Vert a\Vert$ and $\Vert b\Vert$ denote the magnitude of vectors $a$ and $b$ respectively and $\theta$ is the angle between them.

The magnitude of the torque T on the crank when downward force is applied on pedal is $\Vert V\times F\Vert$ , where $V$ is position vector of crank.

Now, magnitude of $V$ is,

  $\begin{array}{c}\Vert V\Vert =\sqrt{{\left(0.16\right)}^2\left({\cos }^{2}30°+{\sin }^{2}30°\right)}\\ =0.16\end{array}$

Now, magnitude of $F$ is,

  $\begin{array}{c}\Vert F\Vert =\sqrt{{\left(2000\right)}^2}\\ =2000\end{array}$

Therefore, magnitude of torque is $T=\Vert V\times F\Vert$ ,

  $\begin{array}{c}\Vert V\times F\Vert =\Vert V\Vert \cdot \Vert F\Vert \cdot \sin \theta \\ =0.16\cdot 2000\cdot \sin 60°\\ =320\left(\frac{\sqrt{3}}{2}\right)\\ =160\sqrt{3}\end{array}$

Thus, the magnitude of torque Tis $\Vert T\Vert =160\sqrt{3}$ .