Determine the magnitude of F1 and its direction θ so that the resultant force is directed vertically upward and has a magnitude of 800 N.

#### Solution:

### We will first draw the vector components as follows.

### The x and y component of each force is also shown, depicted by the color that corresponds to the force itself.

### Now we can write down the x and y components of each force.

### (F_1)_x=F_1\text{sin}\theta\,N

### (F_1)_y=F_1\text{cos}\theta\,N

### (F_2)_x=400\text{ cos 30}^0=346.41\,N

### (F_2)_y=400\text{ sin 30}^0=200\,N

### (F_3)_x=600(\frac{4}{5})=480\,N

### (F_3)_y=600(\frac{3}{5})=360\,N

### We can now sum the x components together and the y components together. To do this, we will first establish which sides we will consider to be positive. In this case, we will choose forces going up and forces going to the right as positive.

### +\rightarrow\sum(F_R)_x=\sum(F_x)

### (F_R)_x=0=F_1\text{sin}\theta+346.41-480

### F_1\text{sin}\theta=133.6

### +\uparrow\sum(F_R)_y=\sum(F_y)

### (F_R)_y=800=F_1\text{cos}\theta+200+360 (Remember in the question, it tells us that resultant is 800 N upwards)

### F_1\text{cos}\theta=240

Hello. Please may you show me How to solve simultaneously

F1sin=133.6

F1cos=240

Thank you very much

To do so, you only need to remember that sin/cos = tan. So you can isolate eq.1 for F1, and then substitute that back into the other equation. You will then get an equation where it’s sin/cos, which is tan, and you can plug it into your calculator. If you need more help, please let me know, thanks! 🙂

Thank you so much you I really understood

I am really glad to hear that!