A 313-kg boat is sailing 15.7 A 313-kg boat is sailing 15.7 degree north of east

Question

A 313-kg boat is sailing 15.7

A 313-kg boat is sailing 15.7 degree north of east at a speed of 1.84 m/s. Thirty seconds later, it is sailing 32.7 degree north of east at a speed of 4.36 m/s. During this time, three forces act on the boat: a 33.3-N force directed 15.7 degree north of east (due to an auxiliary engine), a 22.9-N force directed 15.7 degree south of west (resistance due to the water), and (due to the wind). Find the magnitude and direction of the force Express the direction as an angle with respect to due east.

Explanation / Answer

Mass of the boat m = 313 Kg Initial speed v1 = 1.84 [cos 15.7 i + sin 15.7 j] = 1.7713 i + 0.4979j m/s time t = 30 s Final speed v2 = 4.36 [ cos 32.7 i + sin 32.7 j] = 3.6689 i + 2.355 j m/s

v2 - v1 = ( 3.6689 i + 2.355 j m/s - (1.7713 i + 0.4979) = 1.8976 i + 1.8571 j m/s Acceleration of the boat a = (v2 - v1)/t = (1.8976 i + 1.8571j)/30 = 0.0632534 i + 0.0619034 j m/s^2

Net Force acting on the boat F = ma = 313 * ( 0.0632534 i + 0.0619034j) = 19.79725 i + 19.37 j N Given that F1 = 33.3 [ cos15.7 i + sin15.7 j] = 32.057635156 i +9.010994851 j N

F2 = - 22.9 [cos15.7 i + sin15.7 j] = -22.0456409932i - 6.19675021285 j N Fw = ?

We can write from the given data F = F1 + F2 + Fw Fw = F - F1 - F2

= (19.79725 i + 19.37 j ) - (32.057635156 i +9.010994851 j) - (-22.0456409932i - 6.19675021285 j)

= 9.78585 i + 16.55571N

Magnitude = 19.231598837 N

Direction = Tan^-1[19.23/9.78585]

= 59.4153638208 degrees North of east

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