Determine the magnitude of F1 and its direction 4

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.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

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

The next step is to solve the two equations. Note that you will have to use the identity which says \frac{\text{sin}\theta}{\text{cos}\theta}=\text{tan}\theta

When the equations are simultaneously solved, we get:

\theta=29.1^0

F_1=275\,N

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-50.