# Help, please on vectors and newton's laws

1. Jul 24, 2011

### jehan4141

Hello :) This is my question:

A 325-kg boat is sailing 15 degrees north of east at a speed of 2 m/s. Thirty seconds later, it is sailing 35 degrees north of east at a speed of 4 m/s. During this time, three forces act on the boat: a 31 N force directed 15 degrees north of east (due to an auxiliary engine), a 23.0 N force directed 15 degrees south of west (resistance due to water), and Fw (due to wind). Find the magnitude and direction of the force Fw. Express the direction as an angle with respect to due east.

----------------------------------------------------------------------------------------
To my understanding, the first part of the word problem is irrelevant. After the thirty seconds, the boat sails at a constant 4 m/s. Thus, acceleration is 0. That means the net forces all cancel out to zero.

This is how I calculated it:

Fx = 31cos15 + (-23cos15) + Fwx = 0
Fx = -7.72740661

Fy = 31 sin15 + (-23sin15) + Fwy = 0
Fy = -2.070552361

F = SQRT( Fx^2 + Fy^2)
F = 8 N

This answer is wrong. The answer is 18.4 N, 68 degrees north of east.

Can someone tell me why my reasoning/calculations is/are wrong?

2. Jul 24, 2011

### lewando

Your mistake is that you are thinking that the boat's velocity is constant (from t=30 s and thereafter). It is not. The forces that resulted in the change of velocity from t=0 s to t= 30 s (an indication of acceleration) are still in effect and are unchanged at t=30 s.

3. Jul 24, 2011

### Staff: Mentor

You've calculated the net force due to the water resistance and the engine. You haven't determined the additional force due to the wind required to change the velocity of the boat to its new value.

You might want to think in terms of a change in momentum (Δp). Initially the 325kg boat has a certain velocity, so it has a certain momentum. 30 seconds later (Δt) it has a new velocity, hence a new momentum. Some net force acted on the boat to produce this change in momentum (Δp). How is force related to change in momentum and change in time?

This net force will be be the resultant of the sum of the forces acting due to: the water resistance, the engine, and the wind. The wind is the unknown you're looking for.

4. Jul 25, 2011

### jehan4141

I haven't learned about momentum just yet, but with your guyses help, I managed to figure it out ! thank you :D

5. Jul 25, 2011

### jehan4141

GRRR :/ apparently I still do not understand the problem still...so, the velocity doesn't remain a constant 4 m/s after the first 30 seconds?

gneill: my reasoning was that the net force caused by the engine and water resistance needed to be cancelled out by the wind.

6. Jul 25, 2011

### jehan4141

asfd

7. Jul 25, 2011

### Staff: Mentor

It might or it might not. You're not told one way or the other in the problem statement. All you know is: at 30 seconds after the first velocity reading the new velocity is 4 m/s with the new heading.
There's no reason why these forces should cancel. In fact, since you are told that there is a change in velocity there MUST be unbalanced forces at work (acceleration).

As I mentioned previously, one way to solve this problem is to calculate the change in momentum, Δp. Change in momentum over change in time, Δp/Δt, equals the net force that caused the change. Subtract the given constant forces (engine, water) and you're left with whatever it was that did the deed (in this case you're told it's the wind).

Momentum is vector quantity (as are velocity and force). So write each velocity and force as a vector. Make momentum vectors by multiplying the initial and final velocities by the mass of the boat. Calculate Δp = p2 - p1. Divide by Δt = 30secs. There's your net force.

8. Jul 25, 2011

### jehan4141

4+ hours later i officially give up on this problem. thank you again gneill i haven't learned about momentum yet and should be able to figure out this problem without momentum. :) thank you all

9. Jul 25, 2011

### Staff: Mentor

How about an approach that uses Newton's second law? For that you'll need to determine the acceleration vector. For this problem, the average acceleration can be used, as given by the change in velocity divided by the change in time. Once again you need to start by writing the initial and final velocities as vectors.

10. Jul 25, 2011

### superg33k

I hope this helps:

Initial speed East: V_E1=2cos15
Initial speed North: V_N1=2sin15
Final speed East: V_E2=4cos35
Final speed North: V_N2=4sin35

Acceleration East: A_E=(V_E2 - V_E1)/t
Acceleration North: A_N=(V_N2 - V_N1)/t

Force Wind East: F_E=325*A_E-(31-23)cos15
Force Wind North: F_N=325*A_N-(31-23)sin15