# I'm Stumped

1. Oct 10, 2008

### ^Kris

1. The problem statement, all variables and given/known data

The figure below shows a worker poling a boat-a very efficient mode of transportation-across a shallow lake. He pushes parallel to the length of the light pole, exerting on the bottom of the lake a force of 245 N. Assume the pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of 35.0° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 390 kg.

(a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force.

(b) Model the forces as constant over a short interval of time to find the velocity of the boat 0.450 s after the moment described.

2. Relevant equations
F=ma

3. The attempt at a solution
First of all, I would like to say thank you for everyone's help in advance. I stumbled accross this gem of a site while seeking help on my homework. I think its an awesome site.

I got part (a) as 3681.4737 N . I found this by finding the normal force on the boat n = Mg then finding the push force from the pole on the ground in the y direction and subtracting that from the normal.

I just don't understand what the heck part b is asking. Any help in any sort of direction would be great. I'm pretty sure its something simple, I just cannot think of anything to apply. It says the final answer should be in m/s. Thanks again.

2. Oct 10, 2008

### skatj

I think it means that if F is held constant over a period t, then so would acceleration. Knowing that, you can find final V using simple kinematics, given a and t.

3. Oct 10, 2008

### ^Kris

Ok so I used F=ma to find a. I found the horizontal F by doing 245cos(35) and got 200.6922, and then divided that by the mass of the boat which is 390kg and got .3928m/s^2. I know the acceleration, and the time, but I do not know the initial V, therefore how can I find the Final V?

4. Oct 10, 2008

### Rake-MC

How about $$a = \frac{dv}{dt}$$

5. Oct 10, 2008

### skatj

I only skimmed the problem, but isn't initial v given?

6. Oct 10, 2008

### ^Kris

I missed the part with the .857m/s as the initial, but I tried that and put it into the answer block and it says its wrong. I used V = Vi + at to solve for the final speed, but the online answer is saying "Did you accidentally divide or take the inverse in your calculation?" which i dont think I did and I think I carried out the problem correct. Unless I need to incorporate the 47.5 N of force opposed to the keel?

7. Oct 10, 2008

### ^Kris

Wow, I am an idiot. Thank you all for the help. I had to find the acceleration of the opposing force and subtract that from the acceleration of the boat, and then add that to the initial speed given. Thanks again.

8. Oct 10, 2008

### Rake-MC

Well yes you do, You said you had about 200N forwards, and we're given that there is 47.5 N backwards. Sum the forces up, re-calculate acceleration. Use your constant acceleration formula again.
Provided your 200N is correct then you should be fine.

9. Oct 10, 2008

### Rake-MC

Just re-reading what you've done. 35 degrees is with the vertical (as given in the question).

To find the horizontal you should take a sine, not a cosine.

10. Oct 10, 2008

### ^Kris

That's weird b/c I got the correct answer according to online, so why did the cosine calculation provide the correct answer for me?