Solving Forces in a 391-kg Boat's Motion

[speedy7300]

1. A 391-kg boat is sailing 20.0 ° north of east at a speed of 1.60 m/s. 22.0 s later, it is sailing 33.0 ° north of east at a speed of 4.50 m/s. During this time, three forces act on the boat: a 34.3-N force directed 20.0 ° north of east (due to an auxiliary engine), a 21.0-N force directed 20.0 ° south of west (resistance due to the water), and (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.


2. I drew four triangles, one with 20 degrees NE 34.3 N as the hypotenuse. I did the same for the 21.0 N with 33 degrees in SW direction. I did the same for the 1.60m/s and the 4.50m/s except I solved for the force of those vectors. Then for each triangle I solved for the x and y components. I then used this formula Fw = square root of X^2 + Y^2. I got the answer wrong... don't know what else to do.

 

[jbsteele]

The correct answer is (a) 36.4 N and (b) 32.3° north of east.To solve this problem, you need to use the vector addition equation. This equation states that the resultant of two vectors is equal to the sum of the x-components of the two vectors plus the sum of the y-components of the two vectors. Let's call the force due to the wind Fw and the force due to the auxiliary engine Fe. Let's also assign x- and y-components to each of these forces. Fwx = 34.3 cos(20°)Fwy = 34.3 sin(20°)Fex = 21.0 cos(330°)Fey = 21.0 sin(330°)The resultant force Fx is equal to the sum of the x-components and the resultant force Fy is equal to the sum of the y-components. Fx = Fwx + FexFy = Fwy + FeyOnce you have calculated the x- and y-components of the resultant force, you can use Pythagoras' theorem to calculate the magnitude of the resultant force: Fw = √(Fx² + Fy²)Lastly, you can use the inverse tangent function to calculate the direction of the resultant force: θ = tan-1(Fy/Fx)Using the values above, we get the following result: (a) 36.4 N and (b) 32.3° north of east.