Chapter 2, Problem 1RP

Determine the magnitude of the resultant force F_R and its direction, measured clockwise from the positive u axis.

Prob. R2–1

To determine

Find the magnitude of the resultant force ${\text{F}}_R$.

Find the angle $\varphi$ is measured clockwise from the positive u axis.

Answer to Problem 1RP

The magnitude of the resultant force ${\text{F}}_R$ is $\underset{\_}{605\text{N}}$.

The angle $\varphi$ is measured clockwise from the positive u axis is $\underset{\_}{85.4°}$.

Explanation of Solution

Given information:

The component force $F_1=300\text{N}$.

The component force $F_2=500\text{N}$.

The angle between the $u\text{and}v$ coordinate axis is $\theta =70°$

The angle between the component force $F_1$ and the force in the direction of u axis is $\theta =30°$.

The angle between the component force $F_2$ and the force in the direction of v is $\theta =45°$.

Calculation:

Apply the parallelogram law of addition and triangular rule.

Steps to draw the parallelogram law of addition are as follows:

1. 1. From the head of magnitude of force $F_1$, draw a line parallel to the component force $F_2$.
2. 2. From the head of component of force $F_2$, draw a line parallel to the component of force $F_1$.
3. 3. Join the above two lines at an intersection point to form the adjacent sides of a parallelogram.
4. 4. Draw the resultant force $F_R$ at the intersection point.

Draw the resultant of two forces $F_1\text{and}{F}_2$ using the above steps as in Figure (1).

Using Figure (1),

Write the formula for cosine law to calculate the magnitude of the resultant force $F_2$ using Figure (2).

$F_R=\sqrt{{\left(F_1\right)}^2+{\left(F_2\right)}^2-2F_1{F}_2\cos c}$ (1)

Here, the angle is c.

Calculate the angle c from Figure (1).

$\begin{array}{c}c=\frac{360°-\left[2\left(70°-30°+45°\right)\right]}{2}\\ =95°\end{array}$

Apply the sine rule by using Figure (1) to calculate the angle $\alpha$.

$\frac{\sin \alpha }{F_2}=\frac{\sin c}{F_R}$ (2)

Determine the angle between the resultant force and the u axis.

$\theta =\alpha +30°$ (3)

Conclusion:

Substitute 300 N for $F_1$, 500 N for $F_2$, and $95°$ for c in Equation (1).

$\begin{array}{c}{F}_R=\sqrt{{\left(300\text{N}\right)}^2+{\left(500\text{N}\right)}^2-2\left(300\text{N}\right)\left(500\text{N}\right)\cos 95°}\\ =\sqrt{90,000+250,000-\left(300,000\times -0.087\right)}\\ =605.062\text{N}\\ \approx 605\text{N}\end{array}$

Thus, the magnitude of the resultant force is $\underset{\_}{605\text{N}}$.

Substitute 500 N for $F_2$, 605 N for $F_R$, and $95°$ for c in Equation (2).

$\begin{array}{l}\frac{\sin \alpha }{500\text{N}}=\frac{\sin 95°}{605\text{N}}\\ \frac{\sin \alpha }{500\text{N}}=\frac{0.996}{605\text{N}}\\ \sin \alpha =0.82314\\ \alpha =55.40°\end{array}$

Substitute $55.40°$ for $\theta$ in Equation (3).

$\begin{array}{c}\varphi =55.40+30°\\ =85.40°\end{array}$

Thus, the direction $\varphi$ of magnitude of the resultant force measured clockwise from the positive u axis is $\underset{\_}{85.4°}$.