Chapter 9.1, Problem 71E

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.

71.

To determine

To find: The resultant force of force $F_1$, force $F_2$ and force $F_3$.

Answer to Problem 71E

The Resultant force acting at the given point is $-7.565i+10.61j$.

Explanation of Solution

Given:

Magnitude of Force $F_1$ is 10.

Magnitude of Force $F_2$ is 8.

Magnitude of Force $F_3$ is 6.

Force $F_1$ , force $F_2$ and force $F_3$ are in shown in Figure(1).

Formula used:

If $r$ is the magnitude of a force $F$ and $\theta$ is the angle made by force $F$ with positive direction of $x$-axis, then force is represented as

$F=r\cos \theta i+r\sin \theta j$ (1)

Calculation:

Force vectors $F_1$, force $F_2$ and force $F_3$ with magnitude and angles on coordinate axis are shown in Figure (1).

Figure (1)

Magnitude of $F_1$ is 10 and vector $F_1$ makes $60°$ with positive $x$ -axis direction.

Substitute 10 for $r$ and $60°$ for $\theta$ in equation (1).

$\begin{array}{c}{F}_1=r\cos \theta i+r\sin \theta j\\ =10\cos 60°i+10\sin 60°j\\ =10\times \left(\frac{1}{2}\right)i+10\left(\frac{\sqrt{3}}{2}\right)j\\ =5i+8.66j\end{array}$ (2)

Magnitude of $F_2$ is 8.

Angle made by $F_2$ with positive $x$ -axis is $\left(180°-30°\right)$ or $150°$.

Substitute $8$ for $r$ and $150°$ for $\theta$ in equation (1).

$\begin{array}{c}{F}_2=8\cos 150°i+8\sin 150°j\\ =8\times \left(-0.866\right)i+8\times 0.5j\\ =-6.928i+4j\end{array}$ (3)

Magnitude of $F_3$ is 6.

Angle made by $F_3$ with positive $x$-axis is $200°$.

Substitute 6 for $r$ and $200°$ for $\theta$ in equation (1).

$\begin{array}{c}{F}_3=6\cos 200°i+6\sin 200°j\\ =6\times \left(-0.9396\right)i+6\times \left(-0.3420\right)j\\ =-5.6376i-2.052j\end{array}$ (4)

Resultant force $F_r$ acting at a point is

$F_r=F_1+F_2+F_3$ (5)

Substitute $5i+8.66j$ for $F_1$, $-6.928i+4j$ for $F_2$ and $-5.6376i-2.052j$ for $F_3$ in equation (5).

$\begin{array}{c}{F}_r=F_1+F_2+F_3\\ =\left(5i+8.66j\right)+\left(-6.928i+4j\right)+\left(-5.637i-2.05j\right)\\ =\left(-7.565\right)i+10.61j\end{array}$

Thus, resultant force acting at the given point is $-7.565i+10.61j$

To determine

To find: The additional force required for the force $F_1$, force $F_2$ and $F_3$ to be in equilibrium.

Answer to Problem 71E

The additional force required for the force $F_1$, force $F_2$ and $F_3$ to be in equilibrium, is $7.565i-10.61j$.

Explanation of Solution

Given:

The resultant force acting at the given point is $-7.565i+10.61j$ from section (1).

Property used:

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

$F_a+F_r=0$ (6)

Calculation:

From section (1), value of $F_r$ is $-7.565i+10.61j$.

Substitute $-7.565i+10.61j$ for $F_r$ in equation (6).

$\begin{array}{c}{F}_a=-F_r\\ =-\left(-7.565i+10.61j\right)\\ =7.565i-10.61j\end{array}$

Thus, additional force required is $7.565i-10.61j$.