Chapter 3, Problem 42PQ

The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F₁ = 1.90f, F₂ = f, and F₃ = 1.4f, where f is a constant.

a. Use the coordinate system shown in Figure P3.42 to find $\stackrel{\to }{R}={\stackrel{\to }{F}}_1+{\stackrel{\to }{F}}_2+{\stackrel{\to }{F}}_3$ in component form in terms of f.

b. If R_x = 0.33, what is R_y?

c. Check your result by comparing your answer to that of Problem 6.

FIGURE P3.42

To determine

The component form of $\overrightarrow{R}$ in terms of $f$.

Answer to Problem 42PQ

The component form of $\overrightarrow{R}$ in terms of $f$ is $\underset{\_}{\overrightarrow{R}=-0.166f\stackrel{\wedge }{i}-0.188f\stackrel{\wedge }{j}}$.

Explanation of Solution

$\overrightarrow{R}$ in vector component form can be written as follows using figure P3.42.

  $\overrightarrow{R}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3$ (I)

Here, $\overrightarrow{R}$ is the resultant force, ${\overrightarrow{F}}_1$, ${\overrightarrow{F}}_2$ , and ${\overrightarrow{F}}_3$ are the components of force

The force ${\overrightarrow{F}}_2$ with magnitude $f$ can be resolved into two from figure, $-f\cos 30°\stackrel{\wedge }{i}$ and $f\sin 30°\stackrel{\wedge }{j}$, thus ${\overrightarrow{F}}_2$ will be the sum of both. Similarly ${\overrightarrow{F}}_3$ will be the sum of $1.4f\cos 60°\stackrel{\wedge }{i}$ and $1.4\sin 60°\stackrel{\wedge }{j}$.

Conclusion:

Substitute, $1.90f$ for ${\overrightarrow{F}}_1$, $-f\cos 30°\stackrel{\wedge }{i}+f\sin 30°\stackrel{\wedge }{j}$ for ${\overrightarrow{F}}_2$, and $1.4\sin 60°\stackrel{\wedge }{j}+1.4f\cos 60°\stackrel{\wedge }{i}$ for ${\overrightarrow{F}}_3$ in equation (I).

  $\begin{array}{c}\overrightarrow{R}=-1.90f\stackrel{\wedge }{j}-f\cos 30°\stackrel{\wedge }{i}+f\sin 30°\stackrel{\wedge }{j}+1.4\sin 60°\stackrel{\wedge }{j}+1.4f\cos 60°\stackrel{\wedge }{i}\\ =-0.188f\stackrel{\wedge }{j}-0.166f\stackrel{\wedge }{i}\end{array}$

Therefore, the component form of $\overrightarrow{R}$ in terms of $f$ is $\underset{\_}{\overrightarrow{R}=-0.166f\stackrel{\wedge }{i}-0.188f\stackrel{\wedge }{j}}$.

To determine

The value of $R_y$.

Answer to Problem 42PQ

The value of $R_y$ is $\underset{\_}{0.373}$.

Explanation of Solution

Given that $R_x$ is equal to $0.33$.

But from part (a) it is obtained that $\overrightarrow{R}=-0.166f\stackrel{\wedge }{i}-0.188f\stackrel{\wedge }{j}$.

$\overrightarrow{R}$ can be also resolved into two components, ${\overrightarrow{R}}_x$ and ${\overrightarrow{R}}_y$.

  $\overrightarrow{R}={\overrightarrow{R}}_x+{\overrightarrow{R}}_y$ (II)

Rearrange equation (II) to obtain expression for ${\overrightarrow{R}}_y$.

  ${\overrightarrow{R}}_y=\overrightarrow{R}-{\overrightarrow{R}}_x$                                                                                                             (III)

Compare the two values of $\overrightarrow{R}$ the value of ${\overrightarrow{R}}_x$.

  ${\overrightarrow{R}}_x=-0.166f=0.33$                                                                                             (IV)

Conclusion:

From equation (III) find the value of $f$.

  $\begin{array}{c}-0.166f=0.33\\ f=\frac{0.33}{-0.166}\\ =-1.99\end{array}$

Substitute, $-0.166f\stackrel{\wedge }{i}-0.188f\stackrel{\wedge }{j}$ for $\overrightarrow{R}$, $-0.166f$ for ${\overrightarrow{R}}_x$, and $-1.99$ for $f$ in equation (III).

  $\begin{array}{c}{\overrightarrow{R}}_y=-0.166f\stackrel{\wedge }{i}-0.188f\stackrel{\wedge }{j}-{\overrightarrow{R}}_x-0.166f\stackrel{\wedge }{i}\\ =-0.188\times -1.99\\ =0.373\end{array}$

Therefore, the value of $R_y$ is $\underset{\_}{0.373}$.

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.