Yeah... I\'m lost. A 325kg boat is sailing 18 degrees north of east at a speed o

Question

Yeah... I'm lost.

A 325kg boat is sailing 18 degrees north of east at a speed of 1.90 m/s. Thirty seconds later, it is sailing 38.0 degrees north of east at a speed of 3.90 m/s. During this time three forces act on the boat; a 31.0N force directed 18.0 degrees north of east (due to an auxiliary engine), a 23.0N force directed 18.0 degrees south of west (resistance due to water), and (vector)Fw (due to the wind). Find the magnitude and direction of the force Fw. Express the direction as an angle with respect to due east.

Explanation / Answer

Mass of the boat m = 325 Kg Initial speed v1 = 2 [cos 18 i + sin 18 j]               = 1.9021 i + 0.6180 j   m/s time t = 30 s Final speed v2 = 4 [ cos 38 i + sin 38 j]               = 3.152 i + 2.4626 j   m/s v2 - v1 = (3.152 i + 2.4626 j) - (1.9021 i + 0.6180 j)            = 1.2499 i + 1.8446 j m/s Acceleration of the boat a = (v2 - v1)/t                                        = (1.2499 i + 1.8446 j)/30                                                                            = 0.04166 i + 0.06148 j   m/s^2 Net Force acting on the boat F = ma = 325 * (0.04166 i + 0.06148 j)                                                      = 13.54 i + 19.987 j   N Given that    F1 = 31 [ cos18 i + sin18 j] = 29.482 i + 9.579 j N    F2 = - 23 [cos18 i + sin18 j] = -21.874 i - 7.1073 j   N    Fw = ? We can write from the given data                     F = F1 + F2 + Fw                  Fw = F - F1 - F2                       = (13.54 i + 19.987 j) - (29.482 i + 9.579 j) - (-21.874 i - 7.1073 j)                       = 5.932 i + 17.51 j   N Magnitude = 18.4875 N Direction = Tan^-1[17.51/5.932]               = 71.28 degrees North of east     Mass of the boat m = 325 Kg Initial speed v1 = 2 [cos 18 i + sin 18 j]               = 1.9021 i + 0.6180 j   m/s time t = 30 s Final speed v2 = 4 [ cos 38 i + sin 38 j]               = 3.152 i + 2.4626 j   m/s v2 - v1 = (3.152 i + 2.4626 j) - (1.9021 i + 0.6180 j)            = 1.2499 i + 1.8446 j m/s Acceleration of the boat a = (v2 - v1)/t                                        = (1.2499 i + 1.8446 j)/30                                                                            = 0.04166 i + 0.06148 j   m/s^2 Net Force acting on the boat F = ma = 325 * (0.04166 i + 0.06148 j)                                                      = 13.54 i + 19.987 j   N Given that    F1 = 31 [ cos18 i + sin18 j] = 29.482 i + 9.579 j N    F2 = - 23 [cos18 i + sin18 j] = -21.874 i - 7.1073 j   N    Fw = ? We can write from the given data                     F = F1 + F2 + Fw                  Fw = F - F1 - F2                       = (13.54 i + 19.987 j) - (29.482 i + 9.579 j) - (-21.874 i - 7.1073 j)                       = 5.932 i + 17.51 j   N Magnitude = 18.4875 N Direction = Tan^-1[17.51/5.932]               = 71.28 degrees North of east        

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