# Magnitude of the resultant force is to be 500 N 4

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$