I'm Stumped

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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 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.
 

Answers and Replies

  • #2
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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
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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
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How about [tex] a = \frac{dv}{dt} [/tex]
 
  • #5
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I only skimmed the problem, but isn't initial v given?
 
  • #6
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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
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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
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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
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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
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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?
 

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