Three forces act on the ring. If the resultant force F_R has a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of force F_3.

#### Solution:

### We will first write each of the forces in Cartesian vector notation. This will make our calculations much easier to perform. Let us first look at the resultant force, F_R We can write the resultant force like so:

### F_R=120\left\{\text{cos}\,45^0\text{sin}\,30^0i+\text{cos}\,45^0\text{cos}\,30^0j+\text{sin}\,45^0k\right\} N

### We can further simply this.

### F_R=\left\{42.43i\,+\,73.48j\,+\,84.85k\right\} N

##### (To simplify, we expanded the brackets by multiplying each value inside the brackets by 120. Remember theΒ FOIL method)

### Now, we will write force F_1 in Cartesian vector notation like this:

### F_1=80\left\{\frac{4}{5}i\,+\,\frac{3}{5}k\right\} N

### Again, we will simplify this like so:

### F_1=\left\{64.0i\,+\,48.0k\right\} N

##### (Again, we expanded the brackets like before)

### Next up is force F_2. We can write force F_2 like so:

### F_2=\left\{-110k\right\} N

### Finally, we will write the generic Cartesian vector notation for force F_3 as follows:

### F_3=\left\{F_{3_x}i\,+\,F_{3_y}j\,+\,F_{3_z}k\right\} N

### With that, we can equate the resultant force to the addition of all three other forces.

### F_R=F_1\,+\,F_2\,+\,F_3

### \left\{42.43i\,+\,73.48j\,+\,84.85k\right\}=\left\{(64.0+F_{3_x})i\,+\,F_{3_y}j\,+\,(48.0-110+F_{3_z})k\right\}

### Now, we will solve for i, j, and the k components.

##### Remember, we are just solving for each component, same as when we solve for any other equation with x and y variables. In this case, we will equate i groups to the i components, j groups to the j components, and so forth.

### 64.0+F_{3_x}=42.43

### F_{3_x}=-21.57 N

### F_{3_y}=73.48 N

### 48.0-110+F_{3_z}=84.85

### F_{3_z}=146.85 N

### Now, we will find the magnitude of force F_3. We can calculate the magnitude using the following equation:

### F_3=\sqrt{F^2_{3_x}\,+\,F^2_{3_y}\,+F^2_{3_z}}

### F_3=\sqrt{(-21.57)^2\,+\,(73.48)^2\,+\,(146.85)^2}

### F_3=165.62 N

### Last step is to find the coordinate direction angles for force F_3.

### \text{cos}\,\alpha=\frac{F_{3_x}}{F_3}=\frac{-21.57}{165.62}

### \alpha=\text{cos}^{-1}(\frac{-21.57}{165.62})=97.5^0

### \text{cos}\,\beta=\frac{F_{3_y}}{F_3}=\frac{73.43}{165.62}

### \beta=\text{cos}^{-1}(\frac{73.48}{165.62})=63.7^0

### \text{cos}\,\gamma=\frac{F_{3_y}}{F_3}=\frac{146.85}{165.62}

### \gamma=\text{cos}^{-1}(\frac{146.85}{165.62})=27.5^0

This numerical helped me solve another one.Thanks!

Really glad to hear that! Good luck with your studies π

A calculation mistake at Fr, the Y component should be cos45cos30 j, and not cos45sin30 j

Thanks so much for pointing that out. We have fixed it! π