Question 1200093
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The resultant of two forces p and q is 260N, if p is 80N and the angle between p and q is 150°. Find q
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<pre>
We are given two forces  {{{F[1]}}}  and  {{{F[2]}}}  as vectors.

We are given that the magnitude of  {{{F[1]}}}  is p = 80 N.


We are given that the angle between the vectors  {{{F[1]}}}  and  {{{F[2]}}}  is 150°.

We are given the resultant of forces  {{{F[1]}}}  and  {{{F[2]}}}  is equal to 260 N.



    Apply the parallelogram rule of adding forces.

    You will have a triangle with the sides p = 80N and q and the angle between them 
    of 180° - 150° = 30°.

    The opposite side of this triangle is 260N (the resultant).



Apply the cosine law

    {{{260^2}}} = {{{80^2}}} + {{{q^2}}} - 2*80*q*cos(30°).


Transform to the standard form quadratic equation

    q^2 - 2*80*q*(sqrt(3)/2}}} + {{{80^2))) - {{{260^2}}} = 0

    q^2 - 138.56q - 61200 = 0.


Solve the quadratic equation using the quadratic formula.


You will get two roots  326.2 (rounded) and -187.6 (rounded).


Since we are looking for the magnitude, we select the positive root 326.2.


<U>ANSWER</U>.  The magnitude of q is 326.2 N.
</pre>

Solved.



Surely, you can work with any other precision you want - the methodology allows you to do it.