If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force **F** and its direction θ.

#### Solution:

Let us first draw the vector components out. Note that the resultant force is along the positive y-axis as stated in the question.

Now, let us draw the vector components tail to tail.

Using this diagram, we can figure out **F**. To do so, we will use the law of cosines.

F^2=500^2+700^2-2(500)(700)\cos105^0

F=\sqrt{500^2+700^2-2(500)(700)\cos105^0}

F=959.78N

Now, we will use the law of sines to figure out θ.

\dfrac{\sin(90^0+ \theta)}{700}=\dfrac{\sin105^0}{959.78}\theta=45.2^0

What if it were 550N for the resultant force? What would the angle be?

You only need to substitute 550 instead of 500, and then once you find F, you can use the law of sines to figure out the angle.

Why do you multiply 500 and 700 by -2? when finding “F”

We simply plug it into the law of cosines. Please see here: https://www.mathsisfun.com/algebra/trig-cosine-law.html

Hope that helps 🙂