Chapter 9.1, Problem 68E

(a)

To determine

To find: The resultant force of force ${\text{F}}_1$ , force ${\text{F}}_2$ and force ${\text{F}}_3$.

Answer to Problem 68E

The resultant force acting at point P is $\langle 0,0\rangle$.

Explanation of Solution

Given:

Force ${\text{F}}_1$ is ${\text{F}}_1=\langle 3,-7\rangle$, force ${\text{F}}_2$ is ${\text{F}}_2=\langle 4,-2\rangle$ and ${\text{F}}_3$ is ${\text{F}}_3=\langle -7,9\rangle$.

Formula used:

If ${\text{F}}_1=\langle a_1,a_2\rangle$ , ${\text{F}}_2=\langle b_1,b_2\rangle$ and ${\text{F}}_3=\langle c_1,c_2\rangle$ be three vectors, then resultant $F_r$ is

${\text{F}}_r=\langle a_1+b_1+c_1,a_2+b_2+c_2\rangle$ (1)

Where,

• $a_1$ is horizontal component of ${\text{F}}_1$
• $a_2$ is vertical component of ${\text{F}}_1$
• $b_1$ is horizontal component of ${\text{F}}_2$
• $b_2$ is vertical component of ${\text{F}}_2$
• $c_1$ is horizontal component of ${\text{F}}_3$
• $c_2$ is is vertical component of ${\text{F}}_3$.

Calculation:

Force ${\text{F}}_1$ is ${\text{F}}_1=\langle 3,-7\rangle$, force ${\text{F}}_2$ is ${\text{F}}_2=\langle 4,-2\rangle$ and ${\text{F}}_3$ is ${\text{F}}_3=\langle -7,9\rangle$ .

Substitute 3 for $a_1$, $-7$ for $a_2$, 4 for $b_1$ and $-2$ for $b_2$, $-7$ for $a_3$ and 9 for $a_3$ in equation (1), to obtain the resultant force ${\text{F}}_r$.

$\begin{array}{c}{\text{F}}_r=\langle 3+4+\left(-7\right),\left(-7\right)+\left(-2\right)+9\rangle \\ =\langle 0,0\rangle \end{array}$

Thus, resultant force is ${\text{F}}_r=\underset{\_}{\langle 0,0\rangle }$.

(b)

To determine

To find: The additional force required for the forces to be in equilibrium.

Answer to Problem 68E

The additional force required is $\langle 0,0\rangle$.

Explanation of Solution

Given:

Force ${\text{F}}_1$ is ${\text{F}}_1=\langle 3,-7\rangle$, force ${\text{F}}_2$ is ${\text{F}}_2=\langle 4,-2\rangle$ and ${\text{F}}_3$ is ${\text{F}}_3=\langle -7,9\rangle$

Calculation:

Three forces acting at same point are in equilibrium if the resultant sum of the three forces is zero.

If ${\text{F}}_a$ is the additional force required for equilibrium, and ${\text{F}}_r$ be the resultant of ${\text{F}}_1$, $F_2$ and ${\text{F}}_3$ , then,

${\text{F}}_a+{\text{F}}_r=0$

${\text{F}}_a=-{\text{F}}_r$

From section (1) value of $F_r$ is ${\text{F}}_r=\langle 0,0\rangle$.

To find negative of a vector, multiply components by $-1$.

$\begin{array}{c}{\text{F}}_a=\langle -1\times 0,-1\times 0\rangle \\ =\langle 0,0\rangle \end{array}$

Thus, additional force required is ${\text{F}}_a=\underset{\_}{\langle 0,0\rangle }$.