Calculating Line Strength to Stop a Salmon

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The discussion revolves around calculating the minimum line strength needed to stop a salmon weighing 76 N within a distance of 13 cm, given its initial speed of 3.1 m/s. Participants emphasize the importance of using kinematic equations to find acceleration, specifically noting that the formula a = (vf^2 - vi^2) / (2 * Δx) is applicable. The conversation also highlights the need to convert units to SI for accurate calculations, confirming that the calculated acceleration of 36.96 m/s² is correct. Participants clarify that the weight of the fish does not affect the net force when it is swimming at a constant depth. The thread concludes with appreciation for the assistance in solving the problem.
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All right, I am doing my physics homework and I just cannot figure this one out. I have been trying for literally over an hour, so I am running out of ideas. Here it is.

The tension at which a fishing line snaps is commonly called the line's “strength.” What minimum strength is needed for a line that is to stop a salmon of weight 76 N in 13 cm if the fish is initially drifting at 3.1 m/s? Assume a constant deceleration.

All right, so clearly this has to do with force, which is going to end up being the mass of the fish times its decleration. But because they are asking for strength, I am guessing it a magnitude and not a vector. So, I have tried things such as using the velocity over the meters taken to stop the fish, to get time and then apply that to get the acceleration, but that was clearly wrong as the time changes due to the deceleration. I have done all kinds of things, I need help to figure this one out. Thank you
 
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Use your knowledge of kinematics to find the acceleration. (You are given the initial speed and the distance--that's all you need.) Then use Newton's 2nd law to find the force needed to produce that acceleration. (What's the mass of the fish?)
 
Thank you so much. I hate those problems that are so simple, yet you overlook. I figured out the answer and it was correct. I used the formula 2a(Change in x) = (final velocity) squared minus (initial velocity) squred. but once again, thank you
 
another question about the fish problem

I was wondering if someone could explain the fish problem in a bit more detail, I can't seem to get it...

Thanks in advance
 
mike_24 said:
I was wondering if someone could explain the fish problem in a bit more detail, I can't seem to get it...

Thanks in advance
Welcome to the Forums Mike,

How would you approach the problem yourself? How much do you know about kinematics?
 
As far as approaching the problem, I am not sure, I was doing a few things with F=ma and replaceing the F with W, but i don't seem to be making a whole lot of progress.
 
mike_24 said:
As far as approaching the problem, I am not sure, I was doing a few things with F=ma and replaceing the F with W, but i don't seem to be making a whole lot of progress.
Okay, do you know any kinematic equations?
 
Umm what would be an example of some?
 
Last edited by a moderator:
  • #10
Yes I do know some kinematics
 
  • #11
I am familiar with all of those yes
 
  • #12
mike_24 said:
Yes I do know some kinematics
Make two list, one with variables you know. The other with variable(s) you wish to find. Which formulae are applicable in your case?
 
  • #13
I don't see how the initial speed and the distance can get you the acceleration though.
 
  • #14
mike_24 said:
I don't see how the initial speed and the distance can get you the acceleration though.
You also know the final velocity.
 
  • #15
For this problem the displacement and velocity equation would be needed i think, and Newtons 2nd law F=ma and maybe W=mg
 
  • #16
Ohhh, it would be zero
 
  • #17
mike_24 said:
Ohhh, it would be zero
Spot on.
mike_24 said:
For this problem the displacement and velocity equation would be needed i think,
Good.
mike_24 said:
and Newtons 2nd law F=ma
Sounds good.
mike_24 said:
and maybe W=mg
You may want to rethink this. What can you say about the net vertical force is a fish is floating (or swimming) at the same depth?
 
  • #18
it would zero if the fish was at the same depth?
 
  • #19
mike_24 said:
it would zero if the fish was at the same depth?
Correct, so you don't need to worry about the weight of the fish. However, you do need the weight to calculate the ...
 
  • #20
the mass of the fish?
 
  • #21
mike_24 said:
the mass of the fish?
Correct. Can you have a go at the question from here?
 
  • #22
Yes, i hope so, but one other question, I rearranged the velocity and displacement equation to solve for acceleration and have calculated the mass of the fish, but i get a really large acceleration, is it because i converted the 13cm to m?, isn't that was is always done?, or am I going wrong somewheres else?
 
  • #23
So, once you re-arrange your equation you should obtain;

a = \frac{v_{f}^2}{2\Delta x}

Yes, you should always convert your units to S.I. units in this type of problem.
 
  • #24
Doing that gives you an acceleration of 36.96m/s^2, just seems like it should be more like 3
 
  • #25
Would 36.96 be the correct acceleration?
 
  • #26
mike_24 said:
Would 36.96 be the correct acceleration?
Yes, that's correct. The units are m/s^2.
 
  • #27
As Doc Al said, your answer is correct. However, I would be interested to know why you think the acceleration should be around 3m.s-2?
 
  • #28
I had no reason to believe it would be around 3, it was just it seemed 36m/s^2 would be too high of a number. Thanks though for all the help with this problem, I appreciate it.
 
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