A 292-kg boat is sailing 14.2° north of east at a speed of 1.96 m/s. Thirty seco

Question

A 292-kg boat is sailing 14.2° north of east at a speed of 1.96 m/s. Thirty seconds later, it is sailing 38.1° north of east at a speed of 3.60 m/s. During this time, three forces act on the boat: a 32.7-N force directed 14.2° north of east (due to an auxiliary engine), a 23.9-N force directed 14.2° south of west (resistance due to the water), and W (due to the wind). Find the magnitude and direction of the force W. Express the direction as an angle with respect to due east.
magnitude N
direction °

Explanation / Answer

Mass of the boat m = 292 Kg Initial speed u = 1.96[cos 14.2 i + sin 14.2 j] m/s                       = 1.9 i + 0.48 j m/s Final speed v = 3.6[cos 38.1 i + sin38.1 j] m/s                      = 2.83 i + 2.22 j m/s time taken t = 30 s From kinematic relation                   Acceleration a = (v - u)/t                                          = [(2.83 i + 2.22 j ) - (1.9 i + 0.48 j)]/30                                          = 0.031 i + 0.058 j m/s^2 Net force acting F = ma                              = (292 Kg)(0.031 i + 0.058 j m/s^2)                              = 9.052 i + 16.936 j N Forces acting on the boat F1 = 32.7[ cos14.2 i + sin14.2 j]      = 31.7 i + 8.02 j N F2 = - 23.9 [ cos14.2 i + sin 14.2 j]      = -23.17 i - 5.862 j N wind force W = ? From the given data                 F = F1 + F2 + W So wind force W = F - F1 - F2                            = ( 9.052 i + 16.936 j) - (31.7 i + 8.02 j) - (-23.17 i - 5.862 j)                            = 0.522 i + 14.778 j N Magnitude of wind force W = Sqrt[(0.522)^2 + (14.778)^2] = 14.8 N Direction = Tan^-1[14.778/0.522]                     = 87.970 north of east          Direction = Tan^-1[14.778/0.522]                     = 87.970 north of east

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