Chapter 11.3, Problem 63E

$\left(a\right)$

To determine

To calculate: The magnitude of torque T as function of force $F=-pk$ .

Answer to Problem 63E

The magnitude of torque T as function of force $F=-pk$ is $T=\frac{p}{2}\cos 40°$ .

Explanation of Solution

Given information:

When brakes are applied in a bicycle with help of downward force of p pounds on pedals, the crank which is of six-inch make a $40°$ angle with horizontal, the position vector of the crank is $V=\frac{1}{2}\left(\cos 40°j+\sin 40°k\right)$ and the force is $F=-pk$ .

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 p pounds on pedals, the crank which is of six-inch make a $40°$ angle with horizontal, the position vector of the crank is $V=\frac{1}{2}\left(\cos 40°j+\sin 40°k\right)$ and the force is $F=-pk$ .

Diagram of the above situation is provided below,

  

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

  $\begin{array}{c}\theta =180°-90°-40°\\ =50°\end{array}$

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,

  $\begin{array}{c}\Vert V\Vert =\sqrt{\frac{1}{4}\left({\cos }^{2}40°+{\sin }^{2}40°\right)}\\ =\sqrt{\frac{1}{4}\left(1\right)}\\ =\frac{1}{2}\end{array}$

Now, magnitude of $F$ is,

  $\begin{array}{c}\Vert F\Vert =\sqrt{p^2}\\ =p\end{array}$

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

  $\begin{array}{c}\Vert V\times F\Vert =\Vert V\Vert \cdot \Vert F\Vert \cdot \sin \theta \\ =\frac{p}{2}\sin 50°\\ =\frac{p}{2}\sin \left(90°-50°\right)\\ =\frac{p}{2}\cos 40°\end{array}$

Thus, the magnitude of torque T as function of force $F=-pk$ is $T=\frac{p}{2}\cos 40°$ .

$\left(b\right)$

To determine

To fill: The table “ $\begin{array}{cccccccc}p& 15& 20& 25& 30& 35& 40& 45\\ T& & & & & & & \end{array}$ ” with help of $T=\frac{p}{2}\cos 40°$ .

Answer to Problem 63E

The complete table is $\begin{array}{cccccccc}p& 15& 20& 25& 30& 35& 40& 45\\ T& 5.75& 7.66& 9.58& 11.49& 13.41& 15.32& 17.24\end{array}$ .

Explanation of Solution

Given information:

The table “ $\begin{array}{cccccccc}p& 15& 20& 25& 30& 35& 40& 45\\ T& & & & & & & \end{array}$ ” and the relation $T=\frac{p}{2}\cos 40°$ .

Consider the provided table “ $\begin{array}{cccccccc}p& 15& 20& 25& 30& 35& 40& 45\\ T& & & & & & & \end{array}$ ” and the relation $T=\frac{p}{2}\cos 40°$ .

Substitute the value of p as 15, the above relation,

  $\begin{array}{c}T=\frac{p}{2}\cos 40°\\ T=\frac{15}{2}\cos 40°\\ =5.75\end{array}$

Therefore, value of Tis $5.75$ .

Substitute the value of p as 20, the above relation,

  $\begin{array}{c}T=\frac{p}{2}\cos 40°\\ T=\frac{20}{2}\cos 40°\\ =7.66\end{array}$

Therefore, value of Tis $7.66$ .

Substitute the value of p as 25, the above relation,

  $\begin{array}{c}T=\frac{p}{2}\cos 40°\\ T=\frac{25}{2}\cos 40°\\ =9.58\end{array}$

Therefore, value of Tis $9.58$ .

Similarly, the other values are evaluated.

Thus, the complete table is $\begin{array}{cccccccc}p& 15& 20& 25& 30& 35& 40& 45\\ T& 5.75& 7.66& 9.58& 11.49& 13.41& 15.32& 17.24\end{array}$ .