Determine the magnitude and orientation

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.

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

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 $700\cos 30^0$ .

$1500=700\cos 30^0+F_B\sin \theta$

$F_B\sin \theta=893.8$ —————————(2)

(Again, we isolated for $F_B\sin \theta$ )

Now, we can solve for both θ and $F_B$ by solving both equations (1) and (2) simultaneously.

Solving yields us with:

$\theta =68.6^0$

$F_B=960N$

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-42.

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.

1500=700\cos 30^0+F_B\sin \theta

F_B\sin \theta=893.8 —————————(2)

(Again, we isolated for F_B\sin \theta)

Now, we can solve for both θ and F_B by solving both equations (1) and (2) simultaneously.