Determine the magnitude and orientation of so that the resultant force is directed along the positive y axis and has a magnitude of 1500 N.

#### Solution:

### Let us first draw the vector forces and their respective x and y components.

### Let us also draw the resultant force which is to be directed along the positive y axis as stated in the question.

### Now, let us establish the positive sides. we will assume \rightarrow^+ is positive and \uparrow+ is positive.

### From the resultant force, we know there are no x components to it (because it is simply a force directed along the positive y axis). Therefore, we can say that all the x components of F_1 and F_2 must add up to 0. The x component of F_1 is equal to 700\sin 30^0.

### Thus we can write:

### 0=700\sin 30^0 -F_B\cos \theta

### F_B\cos \theta=350 —————————(1)

##### (we simply isolated for F_B\cos \theta)

### Again, looking at the resultant force, we see that the y component must equal 1500 N. Thus, we can write the following noting that the y component of F_1 is equal to 700\cos 30^0.