Question

A force $\overrightarrow{F_1}$of magnitude 6.00units acts on an object at the origin in a direction $\theta =30.0^{\circ }$ above the positive x axis in Fig. P3.10. A second force $\overrightarrow{F_2}$of magnitude of 5.00units acts on the object in the direction of the positive y axis. Find graphically the magnitude and direction of the resultant force $\overrightarrow{F_1}+\overrightarrow{F_2}$.

Short answer

Magnitude of$F_1⃗+F_2⃗=9.5\hspace{0.33em}N$and its direction is 57⁰above the positive x axis.

Step-by-step solution

Define vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

State given data

Magnitude of $\overrightarrow{F_1}$, 6 units, acts at the origin.

$\theta ={30}^{\circ }$

Magnitude of $\overrightarrow{F_2}$is 5 units and acts in the direction of the positive y axis.

Step 3: Find graphically the magnitude and direction of the resultant force 

$$\begin{gathered} \overrightarrow{F_2}=5.00+(6\times \sin {30}^{\circ }) \\ =5+3j \\ =8j \end{gathered}$$

$\overrightarrow{F}=\sqrt{\overrightarrow{F_1^2}+\overrightarrow{F_2^2}}$

width="148" height="50" role="math">F→=5.1962+8.002=9.5N

$$\begin{gathered} \theta ={\tan }^{-1}\frac{\overrightarrow{F_2}}{\overrightarrow{F_1}} \\ ={\tan }^{-1}\frac{8}{5.196} \\ ={57}^{\circ } \end{gathered}$$

Hence, the magnitude of $\left(F_1\right)⃗+\left(F_2\right)⃗=9.5\hspace{0.33em}N$ and its direction is 57⁰above the positive x axis.