Determine the magnitude of the resultant force acting 1

If F_1 = 600N and Φ=30°, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.

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 out as follows.

Using our vector component diagram, we can start writing down the x and y components of each force (these x and y components are depicted in the dashed lines with the corresponding color to the force). Also note, we set F_1 to 600N and Φ to 30° as stated in the question.

(F_1)_x=600\text{ cos 30}^0=519.62\,N

(F_1)_y=600\text{ sin 30}^0=300\,N

(F_2)_x=500\text{ cos 60}^0=250\,N

(F_2)_y=500\text{ sin 60}^0=433.01\,N

(F_3)_x=450(\frac{3}{5})=270\,N

(F_3)_y=450(\frac{4}{5})=360\,N

To find the resultant, we will sum the x and y forces. To do this, we add the x components together, and we add the y components together, and in this case, we chose up and to the right as the positive side.

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

(F_R)_x=519.62+250-270=499.62\,N

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

(F_R)_y=300-433.01-360=-493.01\,N

Notice we got a negative value for (F_R)_y. This simply means that the value is opposite to that of which we chose to be positive. In other words, the force is actually acting down. Thus, we can say:

(F_R)_y=493.01\,N\downarrow

We can now figure out the magnitude of the resultant force, F_R, using the Pythagorean theorem.

F_R=\sqrt{(F_R)_x^2+(F_R)_y^2}

F_R=\sqrt{(499.62)^2+(493.01)^2}

F_R=701.91\,N

Now, we can figure out θ, and we will measure it clockwise from the x axis as shown in the diagram.

\theta=\text{tan}^{-1}\left[\frac{(F_R)_y}{(F_R)_x}\right]

\theta=\text{tan}^{-1}\left(\frac{493.01}{499.62}\right)

\theta=44.6^0

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-48.