How to Find Force F on a Boat with Given Forces and Motion?

AI Thread Summary
A 325-kg boat is analyzed as it changes direction and speed, with three forces acting on it, including a 31.0N force and a 23.0N force. The goal is to find the unknown force F, which is calculated to be 18.4N at an angle of 68.0° north of east. The discussion emphasizes the importance of understanding the change in velocity to determine acceleration and resultant force. Participants express confusion about the relationship between velocity and force but ultimately clarify their understanding through calculations. The final answer aligns with the expected result, indicating a successful resolution of the problem.
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A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is
sailing 35.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat:
a 31.0N force directed 15.0° north of east, a 23.0N force directed 15.0° south of west and F
Find the force F
Answer : [18.4N, 68.0° north of east]

Fr = F1 + F2 + F3 (due to 3 forces acting on it)

F3 = Fr - F1 - F2 (as question wanted to look for the 3rd force)

F1 = 31cos15° = 30.0N (to the right)

F2 = -(23cos15) = -22.2N (to the right. Or +ve to the left)

a = (v - u)/t

Do I need to search for the vertical, y and the horizontal, x component of the velocity to turn it into resultant Force? Through square root of (Fy^2 +Fx^2)? I do not understand the first 2 sentences of the question. So in the end they will turn into the resultant force acted on the boat? I am confused.

It is a relatively new topic in mechanics to me. I am sorry I could not even make a single line of solution.. >< i have no idea where to start calculating actually. Though I did forces of equilibrium before, this doesn't seems like anything from it.
 
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PandaBoy said:
Im not sure what to do with the velocity provided. Force doesn't involve velocity hmm..
But what about the change in velocity? What might that have to do with force?
 
Oh so I would use the change in velocity to find the acceleration and thus make it into my resultant force? Hmm which means :

Uy = 2sin15° = 0.52m/s Vy = 4sin35° = 2.29m/s
Ux = 2cos15° = 1.93m.s Vx = 4cos35° = 3.28m/s

In which I have t=35s

Which means my a for x and y-component would be :

x : a = (3.28-1.93m/s)/35 = 0.04m/s^2
y : a = (2.29-0.52m/s)/35 = 0.05m/s^2

and from there I find my Fr which is

x : Fr = 13.0N
y : Fr = 16.3N

Then Fr = 20.8N?

Edit : Just tried solving it, I din get the amount as given by the answer. I am not sure where have I gone wrong.
 
PandaBoy said:
In which I have t=35s
Recheck this value for time.
 
Hey I got my answer! Thanks for the help Doc! I understood some part wrongly but I fixed my understanding now.
I think I should have no problem with the other questions now. Appreciate your real direct-to-the-point replies. Have a good day sir!
 
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