Chapter 9.1, Problem 72E

Equilibrium of Forces The forces F₁, F₂, …, F_n acting at the same point P are said to be in equilibrium if the resultant force is zero, that is, if $F_1+F_2+\cdots +F_n=0$. Find (a) the resultant forces acting at P, and (b) the additional force required (if any) for the forces to be in equilibrium.

72.

To determine

To find: The resultant force of force ${\text{F}}_1$ , force ${\text{F}}_2$, force ${\text{F}}_3$ and ${\text{F}}_4$.

Explanation of Solution

Given:

Force ${\text{F}}_1$, force ${\text{F}}_2$, force ${\text{F}}_3$ and force ${\text{F}}_4$ are in figure(1).

Definition used:

If initial point of a vector v is $P\left(x_1,y_1\right)$ and  terminal point is $Q\left(x_2,y_2\right)$

then vector v is,

$v=\langle x_2-x_1,y_2-y_1\rangle$ (1)

Calculation:

Force vectors ${\text{F}}_1$, force ${\text{F}}_2$, force ${\text{F}}_3$ and force ${\text{F}}_4$ with coordinate values are shown in Figure(1).

Figure (1)

Force vector ${\text{F}}_1$ is calculated by equation (1).

Vector $F_1$ is from $\left(2,2\right)$ to $\left(5,1\right)$.

Substitute 2 for $x_1$, 2 for $y_1$, $5$ for $x_2$ and 1 for $y_2$ in equation (1).

${\text{F}}_1=\langle 5-2,1-2\rangle$

${\text{F}}_1=\langle 3,-1\rangle$ (2)

Vector $F_2$ is from $\left(2,2\right)$ to $\left(3,4\right)$.

Substitute 2 for $x_1$, 2 for $y_1$, 3 for $x_2$ and 4 for $y_2$ in equation (1).

${\text{F}}_2=\langle 3-2,4-2\rangle$

${\text{F}}_2=\langle 1,2\rangle$ (3)

Vector $F_3$ is from $\left(2,2\right)$ to $\left(0,1\right)$.

Substitute 2 for $x_1$, 2 for $y_1$, 0 for $x_2$ and 1 for $y_2$ in equation (1).

${\text{F}}_3=\langle 0-2,1-2\rangle$

${\text{F}}_3=\langle -2,-1\rangle$ (4)

Vector $F_4$ is from $\left(2,2\right)$ to $\left(2,-2\right)$.

${\text{F}}_4=\langle 2-2,-2-2\rangle$

${\text{F}}_4=\langle 0,-4\rangle$ (5)

If ${\text{F}}_r$ is the resultant force acting at point $P$,

${\text{F}}_r={\text{F}}_1+{\text{F}}_2+{\text{F}}_3+{\text{F}}_4$

Substitute $\langle 3,-1\rangle$ for ${\text{F}}_1$, $\langle 1,2\rangle$ for ${\text{F}}_{\text{2}}$, $\langle -2,-1\rangle$ for ${\text{F}}_3$, $\langle 0,-4\rangle$ for ${\text{F}}_4$ in above relation.

${\text{F}}_r=\langle 3,-1\rangle +\langle 1,2\rangle +\langle -2,-1\rangle +\langle 0,-4\rangle$

Add all horizontal components together and add all vertical components together.

${\text{F}}_r=\langle 3+1-2+0,-1+2-1-4\rangle$

${\text{F}}_r=\langle 2,-4\rangle$ (6)

Thus resultant vector ${\text{F}}_r$ is $\langle 2,-4\rangle$.

To determine

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

Explanation of Solution

Given:

Force vectors ${\text{F}}_1$, force ${\text{F}}_2$, force ${\text{F}}_3$ and force ${\text{F}}_4$ with coordinate values are shown in Figure(1).

Figure (1)

Calculation:

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

$\begin{array}{c}{\text{F}}_a+{\text{F}}_r=0\\ {\text{F}}_a=-{\text{F}}_r\end{array}$ (7)

From equation (6) value of $F_r$ is $\langle 2,-4\rangle$.

Substitute $\langle 2,-4\rangle$ for ${\text{F}}_r$ in equation (7).

$\begin{array}{c}{\text{F}}_a=-\langle 2,-4\rangle \\ =\langle -1\times 2,-1\times -4\rangle \\ =\langle -2,4\rangle \end{array}$

Thus, additional force required is ${\text{F}}_a=\langle -2,4\rangle$.