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

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Homework Help Overview

The discussion revolves around determining the force acting on a boat given its mass, initial and final velocities, and the forces applied to it. The subject area includes mechanics, specifically the analysis of forces and motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between velocity changes and force, questioning how to incorporate acceleration into their calculations. Some express confusion about the problem's setup and the relevance of the provided velocities.

Discussion Status

Some participants have attempted calculations involving components of velocity and acceleration to find the resultant force. There is a mix of uncertainty and progress, with one participant indicating they have reached an answer after clarifying their understanding.

Contextual Notes

There is a noted confusion regarding the time variable and its application in the calculations. Participants also mention a lack of familiarity with the topic, which may affect their approach to the problem.

PandaBoy
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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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