Chapter 10.3, Problem 65E

(a)

To determine

To find: The expression for torque $T$ on the crank as a function of $p$ if the magnitude of the torque on the crank is $\Vert V\times F\Vert$ .

Answer to Problem 65E

The expression for torque $T$ on the crank as a function of $p$ is $\Vert V\times F\Vert =\frac{p}{2}\cos 40°$ .

Explanation of Solution

Given information:

The position of the crank is $V=\frac{1}{2}\left(\cos 40°j+\sin 40°k\right)$ and the force is $F=-pk$ .

Calculation:

Consider the cross product.

  $\begin{array}{c}V\times F=\left|\begin{array}{ccc}i& j& k\\ 0& \frac{1}{2}\cos 40°& \frac{1}{2}\sin 40°\\ 0& 0& -p\end{array}\right|\\ =\left|\begin{array}{cc}\frac{1}{2}\cos 40°& \frac{1}{2}\sin 40°\\ 0& -p\end{array}\right|i-\left|\begin{array}{cc}0& \frac{1}{2}\sin 40°\\ 0& -p\end{array}\right|j+\left|\begin{array}{cc}0& \frac{1}{2}\cos 40°\\ 0& 0\end{array}\right|k\\ =-\frac{p}{2}\cos 40°i-0j+0k\\ =-\frac{p}{2}\cos 40°i\end{array}$

Find the absolute value.

  $\begin{array}{c}\Vert V\times F\Vert =\left|\left|-\frac{p}{2}\cos 40°i\right|\right|\\ T=\Vert V\times F\Vert =\frac{p}{2}\cos 40°\end{array}$

Therefore, the expression for torque $T$ on the crank as a function of $p$ is $\Vert V\times F\Vert =\frac{p}{2}\cos 40°$ .

(b)

To determine

To complete: The given table.

Answer to Problem 65E

The complete table is shown below.

Explanation of Solution

Given information:

From part (a), the expression for the torque is $T=\Vert V\times F\Vert =\frac{p}{2}\cos 40°$ .

The given table is shown below.

$p$	15	20	25	30	35	40	45
$T$

Calculation:

From part (a), the expression for the torque is $T=\Vert V\times F\Vert =\frac{p}{2}\cos 40°$ . Substitute the values of $p$ for the different value of torque $T$ .

The complete table is shown below.

$p$	15	20	25	30	35	40	45
$T$	5.75	7.66	9.58	11.49	13.41	15.32	17.24

Therefore, the complete table is shown above.