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A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force
against its sails with magnitude of 6 000 N directed toward the northwest (45° N of W). What is the
direction of the resultant acceleration?
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Answer #1
Given these,

Fx = (3000 - F2 cos45 ) i
Fy = F2 sin45 j

==> F = (3000 - F2 cos45 ) i + F2 sin45 j
= (3000 - 6000* cos45 ) i + 6000* sin45 j
= -1242.6 i + 4242.6 j

==> a = [ -1242.6 i + 4242.6 j ] / 2000
= [ -0.6213 i + 2.121 j ]

Therefore the direction of the acceleration is
? =tan-1 ( 2.121 / -0.6213 ) =-73.65o

i.e., 286.35o from the x - axis( i.e., S of W)
answered by: payden hendon
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