**19. ssm A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.0-N force directed 15.0° north of east (due to an auxiliary engine), a 23.0-N force directed 15.0° 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.

So the vector nature of the initial velocity, final velocity and the change in velocity is shown.  Remember acceleration is the change in velocity over a period of time is acceleration.

We need to determine the  components of these vectors:

 

 

 

 

So now we can determine the  components of acceleration

 

 

Ok, now we can find the net force since   We can use the acceleration to find the net force which is the sum of the forces.

 

 

 

So since we want to find FW, we need to solve for that

 

 

We have found the a’s, so we have to find the  components of the forces

 

 

- Sign is due angle being south of west which is in Quadrant III.

 

 

 

 

Now we have  components of the Wind force.  Put them together using Pythagorean theorem and find the Wind force.

 

 

 

 

 

 

 

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This page last updated on January 11, 2020