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.

Determine the magnitude and orientation

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.

Determine the magnitude and orientation solution

Let us also draw the resultant force which is to be directed along the positive y axis as stated in the question.

Determine the magnitude and orientation solution

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.

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