<p>a 325kg boat is sailing 15 degrees north east at a velocity of 2ms^-1. Thirty

Question

<p>a 325kg boat is sailing 15 degrees north east at a velocity of 2ms^-1. Thirty seconds later it is sailing 35 degrees north of east at a velocity of 4ms^-1. During this time three forces act on the boat:a 31N force directed 15 degrees north of east due to an auxillary engine, a 23N force directed 15 degrees south of west due to tidal resistance and the force Fw due tothe wind</p>
<p>find the magnitude and direction of the force Fw</p>
<p>Express the direction as an angle with respect to due east</p>

Explanation / Answer

Initial velocity = 2 m/s at 15° N of E Final velocity = 4 m/s at 35° N of E ? velocity = Force * time = mass * ? velocity We need to draw the velocity vectors and determine the vector that connects the initial velocity vector to the final velocity vector. I hope you have a protractor and ruler, if not draw a good sketch. Draw the + axis and the + y- axis, and the origin at the point of intersection. Label the origin as Point O From the origin, draw 4 cm vector (line) 15° counter-clockwise from the + x-axis. Label the end point as Point A From the origin, draw 8 cm vector (line) 35° counter-clockwise from the + x-axis. Label the end point as Point B. These vectors represent the initial and final velocity vectors. Draw the line from the end point of the initial 15° vector to the end point of the final 35° vector. This line represents the change in velocity. Draw a vertical line down from Point B to a line to a point directly right of Point A. Draw a horizontal right from Point A until it intersects with the vertical line. Label the point of intersection as Point C. You should see the right triangle BAC. The hypotenuse, line AB, of right triangle BAC represents the change in velocity. We need to determine the length of line AC and BC. To determine the lengths, we need to determine the x and y coordinates of Point A and Point B. To do this we need to determine the vertical and horizontal components of each vector. Initial vector Point A Horizontal = 2 * cos 15° Vertical = 2 * sin 15° Point A = (2 cos 15°, 2 sin 15°) Point B Horizontal = 4 * cos 35° Vertical = 4 * sin 35° Point B = (4 * cos 35°, 4 * sin 35°) Length of Line BC = ? y-coordinates = 4 * sin 35° - 2 * sin 15° = 1.78 Length of Line AC = ? x-coordinates = 4 * cos 35° - 2 * cos 15° = 2.34 These velocity changes occurred in 30 seconds. x- acceleration = 1.78 ÷ 30 = 0.0593 m/s^2 y- acceleration = 2.34 ÷ 30 = 0.078 m/s^2 Force = mass * acceleration The x and y components of the force that caused the x and y accelerations = Fx = 325 * 0.0593 = 19.3 N Fy = 325 * 0.078 = 25.35 N The sum of the x components of the 3 forces must = 19.3 N The sum of the y components of the 3 forces must = 25.35 N During this time, three forces act on the boat: a 34.0 N force directed 15.0° north of east (due to an auxiliary engine), a 20.0 N force directed 15.0° south of west (resistance due to the water), and FW (due to the wind). Auxiliary engine 34.0 N @15.0° north of east x = 34 * cos 15 = 32.84 y = 34 * sin15 = 8.8 20 N @ 15.0° south of west x = -20 * cos 15 = -19.3 y = -20 * sin15 = -5.18 Total Force x = 32.84 + -19.3 + Wx = 19.3 Wx = 5.76 N y = 8.8 + -13.4 + Wy = 25.35 Wy = 26.87 N Wind = (5.76^2 + 26.87^2)^0.5 = 27.48 N The tangent of the angle = 25.35 ÷ 5.76 Angle = tan-1 (25.35 ÷ 5.76) = 77.2° Wind force = 27.84N @ 77.2° N of E

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