Chapter 04, Problem 019 A 347-kg boat is sailing 15.00 north of east at a speed

Question

Chapter 04, Problem 019 A 347-kg boat is sailing 15.00 north of east at a speed of 1.90 m/s. 27.0 s later, it is sailing 38.00 north of east at a speed of 3.60 m/s. During this time, three forces act on the boat: a 32.8-N force directed 15.0 degree north of east (due to an auxiliary engine), a 22.4-N force directed 15.0 degree south of west (resistance due to the water), and F (due to the wind). Find the (a) the magnitude and (b) direction of the force . Express the direction as an angle with respect to due east. (a) Number Units (b) Number Units

Explanation / Answer

mass = m = 347 kg

u = 1.9 cos(15 degrees) i + 1.9 sin(15 degrees) j

t = 27 s

v = 3.6 cos(38 degrees) i + 3.6 sin(38 degrees) j

Net force = m (v - u)/t = F1 + F2 + F3

F1 = 32.8 cos(15 degrees) i + 32.8 cos(15 degrees) j

F2 = - 22.4 cos(15 degrees) i - 22.4 sin(15 degrees) j

F3 = m ( v- u ) / t - F1 - F2

F3 = (347/27)*(3.6 cos(38 degrees) i + 3.6 sin(38 degrees) j) - (1.9 cos(15 degrees) i - 1.9 sin(15 degrees) j )

- ( 32.8 cos(15 degrees) i + 32.8 cos(15 degrees) j ) - (- 22.4 cos(15 degrees) i - 22.4 sin(15 degrees) j )

= (28.4846 i + 36.4856 j) - (-0.491756 j+1.83526 i) - (31.6824 j+31.6824 i) + (5.79755 j+21.6367 i)

= (16.6036 i + 11.0925 j)

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Magnitude = sqrt(16.6036^2 + 11.0925^2) = 19.968 N

Direction wrt due to east = arctan(11.0925/16.0036) = 33.75 o

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