Chapter 9.4, Problem 47E

(a)

To determine

The fourth force in terms of unit vectors i, j, and k.

Answer to Problem 47E

The fourth force is $F_4=\left(-\text{7i}-\text{24j+25k}\right)$.

Explanation of Solution

Given:

The magnitude of force $F_1$ is $7\text{i}$, $F_2$ is $24\text{j}$ and $F_3$ is $-25\text{k}$.

Calculation:

The fourth force in vector form is,

$F_4=\left(x\text{i+yj+zk}\right)$ (1)

All four forces are in equilibrium. Therefore,

$F_1+F_2+F_3+F_4=0$ (2)

Substitute $7\text{i}$ for $F_1$, $24\text{j}$ for $F_2$, $-25\text{k}$ for $F_3$ and $\left(x\text{i+yj+zk}\right)$ for $F_4$ in equation (2) to find x, y and z.

$\begin{array}{c}7\text{i+24j-25k+}\left(x\text{i+yj+zk}\right)=0\\ \left(x\text{i+yj+zk}\right)=-7\text{i}-\text{24j+25k}\end{array}$

Compare the coefficient of $\text{i}$, $\text{j}$ and $\text{k}$ on both sides.

$\begin{array}{l}x=-7\\ y=-24\\ z=25\end{array}$

Substitute $-7$ for $x$, $-24$ for $y$ and $25$ for $z$ in equation (1) to find $F_4$.

$F_4=\left(-\text{7i}-\text{24j+25k}\right)$

Thus, the fourth force is $F_4=\left(-\text{7i}-\text{24j+25k}\right)$.

(b)

To determine

The magnitude of the fourth force.

Answer to Problem 47E

The magnitude of the fourth force is $25\sqrt{2}$.

Explanation of Solution

Given:

The magnitude of force $F_1$ is $7\text{i}$, $F_2$ is $24\text{j}$ and $F_3$ is $-25\text{k}$.

Formula to calculate the magnitude of the fourth force is,

$\left|F_4\right|=\sqrt{x^2+y^2+z^2}$ (3)

Where,

• $F_4$ is the fourth force.
• $x$ is the coefficient of the unit vector $\text{i}$.
• $y$ is the coefficient of unit vector $\text{j}$.
• $z$ is the coefficient of unit vector $\text{k}$.

Substitute $-7$ for $x$, $-24$ for $y$ and $25$ for $z$ in equation (3) to find $\left|F_4\right|$.

$\begin{array}{c}\left|F_4\right|=\sqrt{{\left(-7\right)}^2+{\left(-24\right)}^2+{\left(25\right)}^2}\\ =\sqrt{49+576+625}\\ =25\sqrt{2}\end{array}$

Thus, the magnitude of the fourth force is $25\sqrt{2}$.