Chapter 2, Problem 1FP

Determine the magnitude of the resultant force acting on the screw eye and its direction measured clockwise from the x axis.

Prob. F2-1

To determine

The magnitude of resultant force $F_R$ acting on the screw eye and its direction $\theta$ measured clockwise from the $x$ axis.

Answer to Problem 1FP

The magnitude of resultant force $F_R$ and its direction $\theta$ is $\underset{\_}{\text{6}\text{.8}\text{kN}}$ and $\underset{\_}{103°}$ respectively.

Explanation of Solution

Given:

The magnitude of forces are $\text{2}\text{kN}$ and $\text{6}\text{kN}$.

The angle between the x-axis and the force of $\text{2}\text{kN}$ is $45°$.

The angle between the x-axis and the force of $\text{6}\text{kN}$ is $60°$.

Apply the parallelogram law of addition and triangular rule.

Follow the steps to draw the parallelogram law of addition.

1. 1. First, join the tails of the given components at a point to make them concurrent.
2. 2. From the head of magnitude of force $\text{2}\text{kN}$, draw a line, parallel to the magnitude of the force $\text{6}\text{kN}$.
3. 3. From the head of magnitude of force $\text{6}\text{kN}$, draw a line parallel to the magnitude of the force $\text{2}\text{kN}$.
4. 4. Join the above two lines at an intersection point to form the adjacent sides of a parallelogram.
5. 5. Draw the resultant force $F_R$ at the intersection point.

Show the parallelogram law of addition of vectors using the above steps as shown in Figure (1).

Show the triangular rule using Figure (1) as in Figure (2).

Calculate the magnitude of the resultant force $F_R$ from cosine law using Figure (1).

$F_R=\sqrt{{\left(2\text{kN}\right)}^2+{\left(6\text{kN}\right)}^2-2\left(2\text{kN}\right)\left(6\text{kN}\right)\cos \varphi }$ (I)

Write the general formula for sine law.

$\frac{A}{\sin \alpha }=\frac{B}{\sin b}=\frac{C}{\sin c}$ (II)

Here, the sides of the triangle are A, B, and C, and angles are a, b, and c respectively.

Rewrite the Equation (II) using Figure (2) to calculate the direction of the resultant force.

$\frac{\sin \theta }{6\text{kN}}=\frac{\sin \varphi }{F_R}$ (III)

Conclusion:

Calculate the angle $\varphi$ using Figure (2).

$\begin{array}{c}\varphi =180°-\left(90°-45°\right)-30°\\ =180°-45°-30°\\ =105°\end{array}$

Substitute $\text{105}°$ for $\varphi$ in Equation (I).

$\begin{array}{c}{F}_R=\sqrt{{\left(2\text{kN}\right)}^2+{\left(6\text{kN}\right)}^2-2\left(2\text{kN}\right)\left(6\text{kN}\right)\left(\cos 105°\right)}\\ =\sqrt{4+36-\left[24\left(-0.2588\right)\right]}\\ =\sqrt{4+36+6.2112}\end{array}$

$\begin{array}{c}=\sqrt{46.2112}\\ =6.8\text{kN}\end{array}$

Thus, the magnitude of the resultant force $F_R$ is $\underset{\_}{\text{5}\text{.8}\text{kN}}$.

Substitute $\text{6}\text{.8 kN}$ for $F_R$, and $105°$ for $\varphi$ in Equation (III).

$\begin{array}{l}\frac{\sin \theta }{6\text{kN}}=\frac{\sin 105°}{6.8\text{kN}}\\ \frac{\sin \theta }{6\text{kN}}=\frac{0.9659}{6.8\text{kN}}\end{array}$

$\begin{array}{l}\sin \theta =0.8522\\ \theta ={\sin }^{-1}\left(0.8522\right)\\ \theta =58.4°\end{array}$

Hence, the direction $\theta$ of the resultant force is $\underset{\_}{58.4°}$.