Calculating Force to Propel Boat Across Shallow Lake

[^Kris]

Homework Statement

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.

Homework Equations

F=ma

The Attempt at a Solution

First of all, I would like to say thank you for everyone's help in advance. I stumbled across 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.

 

[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.

 

[^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?

 

[Rake-MC]

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

 

[skatj]

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

 

[^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 don't 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?

 

[^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.

 

[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.

 

[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.

 

[^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?