Chapter 3.3, Problem 38E

(a)

To determine

To find: The rate of change of F with respect to $\theta$. That is,$\frac{dF}{d\theta }$.

Answer to Problem 38E

The rate of change of F with respect to $\theta$ is $\underset{\_}{\frac{dF}{d\theta }=\frac{\mu W\left(\sin \theta -\mu \cos \theta \right)}{{\left(\mu \sin \theta +\cos \theta \right)}^2}}$.

Explanation of Solution

Given:

The magnitute of the force is $F=\frac{\mu W}{\mu \sin \theta +\cos \theta }$,

Where, $\mu$ is a costant and W is weight.

Derivative Rule: Quotient Rule

If $f\left(\theta \right)$. and $g\left(\theta \right)$ are both differentiable function, then

$\frac{d}{d\theta }\left[\frac{f\left(\theta \right)}{g\left(\theta \right)}\right]=\frac{g\left(\theta \right)\frac{d}{d\theta }\left[f\left(\theta \right)\right]-f\left(\theta \right)\frac{d}{d\theta }\left[g\left(\theta \right)\right]}{{\left[g\left(\theta \right)\right]}^2}$ (1)

Calculation:

Obtain the derivative of F.

$\begin{array}{c}\frac{dF}{d\theta }=\frac{d}{d\theta }\left(F\right)\\ =\frac{d}{d\theta }\left(\frac{\mu W}{\mu \sin \theta +\cos \theta }\right)\\ \end{array}$

Apply Quotient Rule as shown equation (1),

$\begin{array}{c}\frac{dF}{d\theta }=\frac{\left(\mu \sin \theta +\cos \theta \right)\cdot \frac{d}{d\theta }\left[\mu W\right]-\mu W\cdot \frac{d}{d\theta }\left[\mu \sin \theta +\cos \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}\\ =\frac{\left(\mu \sin \theta +\cos \theta \right)\left[0\right]-\mu W\left[\mu \cos \theta -\sin \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}\\ =\frac{0-\mu W\left[\mu \cos \theta -\sin \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}\\ =\frac{\mu W\left[\sin \theta -\mu \cos \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}\end{array}$

Therefore, the rate of change of F with respect to $\theta$ is $\underset{\_}{\frac{dF}{d\theta }=\frac{\mu W\left[\sin \theta -\mu \cos \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}}$.

(b)

To determine

To find: The value of $\theta$ if the rate of change equal to zero.

Answer to Problem 38E

The rate of change equal to zero when $\underset{\_}{\theta =\arctan \mu }$.

Explanation of Solution

Given:

From part (a), the rate of change of F with respect to $\theta$ is $\frac{dF}{d\theta }=\frac{\mu W\left[\sin \theta -\mu \cos \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}$.

Calculation:

The rate of change equal to zero, $\frac{dF}{d\theta }=0$.

$\begin{array}{l}\frac{\mu W\left[\sin \theta -\mu \cos \theta \right]}{{\left[\mu \sin \theta +\cos \theta \right]}^2}=0\\ \mu W\left[\sin \theta -\mu \cos \theta \right]=0\end{array}$

Since $\mu W\ne 0$, $\sin \theta -\mu \cos \theta =0$.

$\begin{array}{l}\sin \theta =\mu \cos \theta \\ \frac{\sin \theta }{\cos \theta }=\mu \\ \tan \theta =\mu \\ \theta =\arctan \left(\mu \right)\end{array}$

Therefore, the rate of change equal to zero when $\underset{\_}{\theta =\arctan \mu }$.

(c)

To determine

To draw:. The graph of F as a function of $\theta$ and use it to locate the value of $\theta$ for which $\frac{dF}{d\theta }=0$.

Explanation of Solution

Given:

The magnitute of the force is $F=\frac{\mu W}{\mu \sin \theta +\cos \theta }$, $W=50lb$ and $\mu =0.6$.

Use online graphing calculator to draw the graph of F as a function of $\theta$ as shown in Figure 1.

The value of $\theta$ is appoximately 0.54.

Calculation:

Obtain the magnitute of the force F as a function of $\theta$.

$F=\frac{\mu W}{\mu \sin \theta +\cos \theta }$ (2)

Substitute $W=50lb$ and $\mu =0.6$ in equation (2),

$\begin{array}{c}F=\frac{\left(0.6\right)\left(50\right)}{\left(0.6\right)\sin \theta +\cos \theta }\\ =\frac{30}{\left(0.6\right)\sin \theta +\cos \theta }\end{array}$

Therefore, the magnitute of the force F as a function of $\theta$ is $F=\frac{30}{\left(0.6\right)\sin \theta +\cos \theta }$.

From part (b), the rate of change equal to zero when $\theta =\arctan \mu$.

Substitute $\mu =0.6$ in equation $\theta =\arctan \mu$,

$\begin{array}{c}\theta =\arctan 0.6\\ \approx 0.54\end{array}$

Therefore, the value of $\theta$ is appoximately 0.54.

Substitute $\theta \approx 0.54$ in s and obtain the value of F.

$\begin{array}{c}F=\frac{30}{\left(0.6\right)\sin \left(0.54\right)+\cos \left(0.54\right)}\\ =\frac{30}{\left(0.6\right)\left(\text{0}\text{.51414}\right)+\left(\text{0}\text{.85771}\right)}\\ =\frac{30}{\text{1}\text{.166194}}\\ \approx \text{25}\text{.72}\end{array}$

Therefore, the point $\left(\theta ,F\right)\approx \left(0.54,25.72\right)$

Use online graphing calculator and draw the graph of F as a function of $\theta$ as shown below in Figure 1.

From Figure 1, it is observed that the slope of tangent is horizontanl to the curve $F=\frac{30}{\left(0.6\right)\sin \theta +\cos \theta }$ at the point $\left(0.54,25.72\right)$.

Hence, the value of $\theta$ is consistent.