Chapter 6, Problem 76RE

Solution

To find: The direction and magnitude of the resultant force.

The magnitude of the resultant vector is about $412\text{Ib}$ and the direction angle is about $2.07°$ .

Given information:

It is given that a force of $120\text{lb}$ acts on an object at an angle of ${20}^{°}$ and the second force of $300\text{lb}$ acts on the object at an angle of $-5^{°}$ .

Formula used:

The resultant vector of two vectors $\overrightarrow{X}〈u_1,v_1〉$ and $\overrightarrow{Y}〈u_2,v_2〉$ is given by:

  $\overrightarrow{X}+\overrightarrow{Y}=〈u_1+u_2,v_1+v_2〉$

If a vector has a direction angle $\theta$ and magnitude , the components of can be computed using the formula:

  

The magnitude of is the length of the vector , which is found using the distance formula:

  

Calculation:

The magnitude of the first force is $120\text{lb}$ and the direction angle is ${20}^{°}$ .

Find the component form of the first force.

  

The magnitude of the second force is $300\text{lb}$ and the direction angle is $-5^{°}$ .

Find the component form of the second force.

  

Add the vectors and to get the resultant vector .

  

Find the magnitude of the resultant vector.

  $\begin{array}{c}\left|z\right|=\sqrt{{\left(412\right)}^2+{\left(15\right)}^2}\\ =\sqrt{169744+225}\\ =\sqrt{169969}\\ =412\end{array}$

The vertical component of the resultant vector $〈412,15〉$ is equivalent to .

  $\begin{array}{c}15=412\sin \theta \\ \sin \theta =\frac{15}{412}\\ \theta \approx {2.07}^{\circ }\end{array}$

Thus, the magnitude of the resultant vector is about $412\text{Ib}$ and the direction angle is about $2.07°$ .