Chapter 3.4, Problem 7FP

Determine the magnitude of forces F₁, F₂, F₃, so that the particle is held in equilibrium.

To determine

The magnitude of forces $F_1$, $F_2$, and $F_3$ for equilibrium.

Answer to Problem 7FP

The magnitude of force $F_1$ is $\underset{\_}{466\text{N}}$.

The magnitude of force $F_2$ is $\underset{\_}{879\text{N}}$.

The magnitude of force $F_3$ is $\underset{\_}{776\text{N}}$.

Explanation of Solution

Given information:

The given force values are 600 N and 900 N.

Explanation:

Show the free body diagram of the forces acting on the particle as in Figure 1.

Using Figure (1),

Determine the magnitude of forces using equation of equilibrium.

Force along x direction:

$\begin{array}{l}\left[\left(\frac{3}{5}\right)F_3\right]\left(\frac{3}{5}\right)+600-F_2=0\\ 0.36F_3-F_2=-600\end{array}$ (I)

Force along y direction:

$\begin{array}{l}\left(\frac{4}{5}\right)F_1-\left[\frac{3}{5}{F}_3\right]\left(\frac{4}{5}\right)=0\\ 0.8F_1-0.48F_3=0\\ 0.8F_1=0.48F_3\\ F_1=\frac{0.48}{0.8}{F}_3\end{array}$

$F_1=0.6F_3$ (II)

Force along z direction:

$\begin{array}{l}\left(\frac{4}{5}\right)F_3+\left(\frac{3}{5}\right)F_1-900=0\\ 0.8F_3+0.6F_1=900\end{array}$ (III)

Conclusion:

Substitute $0.6F_3$ for $F_1$ in Equation (III).

$\begin{array}{l}0.8F_3+0.6\left(0.6F_3\right)=900\\ F_3\left(0.8+0.36\right)=900\\ F_3=\frac{900}{1.16}\\ F_3=776\text{N}\end{array}$

Thus, the magnitude of force $F_3$ is $\underset{\_}{776\text{N}}$.

Substitute $776\text{N}$ for $F_3$ in Equation (II).

$\begin{array}{c}{F}_1=0.6\times 776\\ =466\text{N}\end{array}$

Thus, the magnitude of force $F_1$ is $\underset{\_}{466\text{N}}$.

Substitute $776\text{N}$ for $F_3$ in Equation (I).

$\begin{array}{l}0.36\left(776\right)-F_2=-600\\ 279.36+600=F_2\\ F_2=879\text{N}\end{array}$

Thus, the magnitude of force $F_2$ is $\underset{\_}{879\text{N}}$.