Question

Three forces are applied to an object, as shown in the figure. Force has a magnitude of 21.3 newtons (N) and is directed 30.0° to the left of the +y axis. Force has a magnitude of 12.0 N and points along the +x axis. A third force is applied such that the vector sum of the three forces is 0 N. What are (a) the magnitude and (b) direction of this third force ? Specify the direction as a positive angle relative to the negative x axis, as shown in the figure.

Answer 1

Solve. Three forces are applied to an object, as shown in the figure. Force has a magnitude of 21.3 newtons (N) and is directed $30.0{}^{\circ}$ to the left of the $+\mathrm{y}$ axis. Force has a magnitude of 12.0 N and points along the $+\mathrm{x}$ axis. A third force is applied such that the vector sum of the three forces is 0 N. What are (a) the magnitude and (b) direction of this third force? Specify the direction as a positive angle relative to the negative x axis, as shown in the figure.

Solution.

A third force is applied such that the vector sum of the three forces is $0\mathrm{N}$ . So we can state:

$$\overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} + \overrightarrow {F _ {3}} = 0$$

First, project onto the $X$ axis:

$$- F _ {1} \cdot \sin 3 0 {}^ {\circ} + F _ {2} - F _ {3} \cdot \cos \varphi = 0$$

Then project onto the $Y$ axis:

$$F _ {1} \cdot \cos 3 0 {}^ {\circ} - F _ {3} \cdot \sin \varphi = 0$$

So we have the system of two equations with two unknown:

$$\left\{ \begin{array}{c} - F _ {1} \cdot \sin 3 0 {}^ {\circ} + F _ {2} - F _ {3} \cdot \cos \varphi = 0 \\ F _ {1} \cdot \cos 3 0 {}^ {\circ} - F _ {3} \cdot \sin \varphi = 0 \end{array} \right.$$

Rewrite:

$$\left\{ \begin{array}{c} F _ {3} \cdot \cos \varphi = - F _ {1} \cdot \sin 3 0 {}^ {\circ} + F _ {2} \\ F _ {3} \cdot \sin \varphi = F _ {1} \cdot \cos 3 0 {}^ {\circ} \end{array} \right.$$

Divide second equation by first:

$$\left\{ \begin{array}{c} \tan \varphi = \frac {F _ {1} \cos 3 0 {}^ {\circ}}{F _ {2} - F _ {1} \sin 3 0 {}^ {\circ}} \\ F _ {3} = \frac {F _ {1} \cos 3 0 {}^ {\circ}}{\sin \varphi} \end{array} \right.$$ $$\left\{ \begin{array}{c} \tan \varphi = \frac {2 1 . 3 \cdot \frac {\sqrt {3}}{2}}{- 2 1 . 3 \cdot \frac {1}{2} + 1 2} = 1 3. 1 \\ F _ {3} = \frac {2 1 . 3 \cdot \frac {\sqrt {3}}{2}}{\sin (\arctan 1 3 . 1)} \approx 1 8. 4 \end{array} \right.$$ $$\left\{ \begin{array}{c} \varphi = \arctan 0. 8 1 4 5 = 8 5. 6 {}^ {\circ} \\ F _ {3} = \frac {1 8 . 4 5}{\sin (8 5 . 6 {}^ {\circ})} \end{array} \right.$$

**Answer:**

the magnitude of the third force is

$$F _ {3} = 2 6. 4 N$$

positive angle relative to the negative $X$ axis is

$$\varphi = 8 5. 6 {}^ {\circ}$$