# Magnitude of the resultant force FR = F1 + F2 8

Determine the magnitude of the resultant force $F_{R} = F_{1} + F_{2}$ and its direction, measured clockwise from the positive u axis.

#### Solution:

Let us draw the vector components. When doing this question, it is highly recommended to draw a very large diagram, otherwise, the v-axis can cause confusion as it might appear to be the resultant force.

Now, we will use the law of cosines to figure out $F_R$. (Forgot the Law of Cosines?)

$(F_{R})^2=300^2+500^2-2(300)(500)\cos95^0$

$F_R=\sqrt{300^2+500^2-2(300)(500)\cos95^0}$

$F_R=605.1N$

To figure out $\theta$ we will use the law of sines. (Forgot the Law of Sines?)

$\dfrac{605.1}{\sin95^0}=\dfrac{500}{\sin \theta}$

$\theta = 55.40^0$

As the question asks us for the direction measured from the positive u-axis, we need to add $30^0$ to our $\theta$ value. Remember that $\theta$ only represents the angle between force $F_1$ and $F_R$.

$\phi=55.40^0+30^0$

$\phi=85.4^0$

## 8 thoughts on “Magnitude of the resultant force FR = F1 + F2”

• questionsolutions Post author

Looking at force F1 and F2, notice how it creates an angle of 45°+40°=85° angle (between the two). Now, notice the dashed line coming from the end point of force F1 heading towards the end point of force FR. That dashed line, and force F2 creates an interior angle (please refer to: https://goo.gl/vBHDxy), where it must equal 180°. Thus, the inside angle must then be 180°-85°=95°. Hope that helps you out! 🙂

• questionsolutions Post author

The reason we add 30° is because the question asks us to find the direction measured clockwise from the positive u axis. Therefore, the resultant force is θ+ 30°, where 30° is the angle between the positive u-axis and the force F1. Remember that θ represents the angle between force F1 and FR, not the total angle measured from the positive u-axis. Hope that helps! 🙂

• questionsolutions Post author