Start by breaking the problem in two - consider the components of the forces in the horizontal direction seprarately from the components of the forces in the vertical direction. You want to have
and
. The 600N force that goes up to the left has a horizontal component of 600N * (-4/5) (note that its sign is negative as it goes to the left), and a vertical component of 600N * 3/5. The 400N force going up to the right has a horizontal component of 400N * cos(30), and a vertical componenet of 400* sin(30). The unknown force
has a horizontal component of
 )
and a vertical component of
 )
. So summing all the horizontal forces and setting that equal to 0 yields:
 + F_1 cos( \theta) + 400N * cos(30) = 0 <br />
)
From this you can determine the value for
)
.
Then sum the forces in the vertical direction:
 + 400*sin(30) = 800<br />
)
This gives you
 )
. The value of
can then be found from:
)^2 + (F_1 sin ( \theta ))^2 }<br />
)
and the direction of
is:
 } {F_1 cos( \theta ) }<br />
)
Linear Mode
