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<p>A classmate and I were trying to figure out this problem to no avail. We realized that ther is no force acting in the "x" direction, so after finding the components for each force vector and then summing up all the x-components and setting them to zero we have one equation with one unknown.</p>
<p>After some manipulation we ended up with: 3cos(&#945; + 35) = cos(&#945;), but we are not sure how to solve for &#945;...</p>
<p>We would appreciate some help on this one. Thanks.</p>
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Explanation / Answer

For resultant of forces to be along vertical direction:

F x = 0

Now force in negative X direction = - F3.cos i = - 160 . cos i

And force in positive X direction = F2.cos (35 + ) + F1.cos i

   = 60.cos(35+ ) + 140. cos i

Fx = 60.cos(35+ ) + 140. cos - 160 . cos = 0

    60. cos (35 + ) = 20 .cos

   3 x (cos 35.cos - sin35.sin) = cos (eaxpansion of cos (a +b))

   2.457 cos - 1.72 sin = cos

   1.457. cos = 1.72 sin

   tan = 0.847    = tan-1 (0.847)

   = 40.26

on solving we get :

= 40.26 0

b) resultant of force = Fy

   = F3.sin + F2 .sin (35 + ) + F1.sin j

= 160 x sin 40.26 + 60 x sin (35 + 40.26) + 140 x sin 40.26

= 251.90 lb along positive Y axis

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