Magnitude and direction of Wind Force

  • Thread starter tryton
  • Start date
  • #1
4
0

Homework Statement


1. Homework Statement
A 373-kg boat is sailing 13.0 ° north of east at a speed of 2.00 m/s. 32.0 s later, it is sailing 33.0 ° north of east at a speed of 3.90 m/s. During this time, three forces act on the boat: a 33.1-N force directed 13.0 ° north of east (due to an auxiliary engine), a 22.9-N force directed 13.0 ° south of west (resistance due to the water), and Fw(due to the wind). Find the (a) the magnitude and (b) direction of the force Fw. Express the direction as an angle with respect to due east.

Homework Equations


Force=ma
Kinematic equations


The Attempt at a Solution


33.1N-22.9N=10.2N Narrowed down the forces as these two are facing the same direction
M=373Kg
V0=2.0m/s
vx0=1.95m/s
vy0=.450m/s
v=3.9m/s
vx=3.27m/s
vy=2.13m/s
theta1=13degrees
theta2=33degrees

I am not sure where to go from here. What is missing is x,y,a, and Force of wind.
Can get x,y,a from breaking Force down into the x and y components and then use that to find F=ma for x=v0t+1/2at^2 but then I don't know if that is right and still don't know how to get the force of the wind...

Is that the right direction to go in at least?
Thanks for any help on this one.
 
Last edited:

Answers and Replies

  • #2
318
0
I think you are supposed to find the resultant force that the boat makes with the auxiliary engine and the water. Do you know how to add components and find the resultant vector?
 
  • #3
4
0
I realized that part of the question didn't show up. I need to find the Fw, the Force of the wind.

I can add the two forces but I don't know how to separate out Fw.

so
Fa=Force auxiliary engine
Fax=22.1cos13=34.25
Fay=33.1sin13=7.45

Fwc=Force of watercurrent and wind
Fwcx=-29.9cos13=-29.1
Fwcy=-29.9sin13=-6.73

Added the Net Force is
Fx=5.12
Fy=.720

Then
F=ma
ax=5.12/373=.0137
ay=.720/373=.00193

a= square root of (.0137 ^2+.00193^2)=.0139

arctan(.00193/.0137)=8.006degrees

But then that is both water AND wind ...
and they just want the force of the wind.

x=V0t+1/2at^2
x= 2.0m/s * 32.0s + 2 * .0139m/s^2 * 32.0s^2
x=92.47m

Do I need to find Y as well? and then add x and y?
 
  • #4
318
0
oh im sorry. I didnt see the Fw part. Well, the next question is, what do you think, conceptually, the net force is going to be, if there is a net velocity of 3.9m/s during that instentaneous time at 32 seconds?
 

Related Threads on Magnitude and direction of Wind Force

  • Last Post
Replies
5
Views
2K
Replies
2
Views
5K
Replies
2
Views
3K
Replies
4
Views
5K
  • Last Post
Replies
13
Views
4K
Replies
1
Views
3K
Replies
5
Views
85K
Replies
3
Views
10K
Replies
1
Views
2K
Replies
1
Views
483
Top