A 2000 kg sailboat experiences an eastward force of 3000 N by the ocean tide and

Question

A 2000 kg sailboat experiences an eastward force of 3000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest ( 45 degress N of W.) what is the direction of the resultant acceleration?

Explanation / Answer

Given that       F1 = 3000 N ( east) F2 = 6000 N     (N ofW) m = 2000 kg Therefore    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)

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