Chapter 9.1, Problem 70E

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.

70. F₁ = i − j, F₂ = i + j, F₃ = −2i + j

To determine

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

Answer to Problem 70E

Resultant force acting at point P is $j$.

Explanation of Solution

Given:

Force ${\text{F}}_1$ is ${\text{F}}_1=i-j$, force ${\text{F}}_2$ is ${\text{F}}_2=i+j$ and ${\text{F}}_3$ is ${\text{F}}_3=-2i+j$.

Formula used:

If $F_r$ is the resultant of forces ${\text{F}}_1$, ${\text{F}}_2$, and ${\text{F}}_3$ then

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

Calculation:

Force ${\text{F}}_1$ is ${\text{F}}_1=i-j$, force ${\text{F}}_2$ is ${\text{F}}_2=i+j$ and ${\text{F}}_3$ is ${\text{F}}_3=-2i+j$.

If $F_r$ is the resultant of forces ${\text{F}}_1$, ${\text{F}}_2$, and ${\text{F}}_3$ then

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

${\text{F}}_r=\left(i-j\right)+\left(i+j\right)+\left(-2i+j\right)$

Combine terms with i, and and terms with j,

$\begin{array}{c}{\text{F}}_r=\left(i+i-2i\right)+\left(-j+j+j\right)\\ =j\end{array}$

Therefore resultant force acting at point P is $j$.

To determine

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

Answer to Problem 70E

Additional force required is ${\text{F}}_a=-j$.

Explanation of Solution

Given:

Force ${\text{F}}_1$ is ${\text{F}}_1=i-j$, force ${\text{F}}_2$ is ${\text{F}}_2=i+j$ and ${\text{F}}_3$ is ${\text{F}}_3=-2i+j$.

Formula used:

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

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

Calculation:

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

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

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

From section (1) value of $F_r$ is ${\text{F}}_r=j$.

$\begin{array}{c}{\text{F}}_a=-\left(j\right)\\ =-j\end{array}$

Thus, additional force required is ${\text{F}}_a=-j$.