A sailboat with a small outboard motor is initially moving 2.00 m/s at a heading

Question

A sailboat with a small outboard motor is initially moving 2.00 m/s at a heading of 15.0O north of east. Half a minute later, it's heading 35.0O north of east at 4.00 m/s. During this time interval, three forces are acting on the 325.0 kg boat (horizontally, parallel to the plane of the water.) The motor is pushing with a 31.0 N force, 15.0O north of east. There is a 23.0 N resistive force directed 15.0O south of west. And there is the force of the wind, Fw. Find the magnitude and direction of this force.

Explanation / Answer

The horizontal net force without wind would be: Fnet = 31.0 - 23.0 = 8.0 N and the acceletarion without wind would be: a = Fnet / mass = 8 / 325 = 0.0246 m/s² Boat speed without wind would be: V = Vo t + (1/2) a t² = 2 · 30 + (1/2) 0.0246 (30)² = 60 + 11.07 = 71.07 m/s. Now, given that heading is changed from 15º to 35º and speed is 4 m/s, we have: 71.07 cos20 - Vs cosD = 4 71.07 sin20 = Vs sinD where Vs is the speed that boat would reach by the action of the wind force and D is the shortest angle of wind direction with the 35.0º heading. Vs = 71.07 sin20 / sinD 71.07 cos20 - (71.07 sin20 / sinD) cosD = 4 71.07 cos20 - 71.07 sin20 cotD = 4 66.78 - 24.31 cotD = 4 62.78 = 24.31 cotD cotD = 2.58 D = 21.16º (or D = 90 + (90 - 35) + 21.16 = 166.16º south of west) Now: Vs = 71.07 sin20 / sin21.16 = 67.32 m/s and finally: Fw = m Vs / t = 325 · 67.32 / 30 = 729.31 N such strong force means a very big sails.

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