Finding the Minimum Strength of a Fishing Line to Stop a Drifting Fish

[j_confused]

Alright, this one seems too easy to be hard! But, I've spent a lot of time on it and I'm stuck, sad to say. Maybe someone can help please??

What min strength is needed for a line that is to stop a fish of 85N in 0.112m if the fish is initially drifting at 2.5 m/s? (Assume constant deacceleration.)

I'm having problems figuring out how to relate velocity to net x and y forces. I'm assuming tension is opposite of fish weight, but aside from that, I don't see any other forces??

Thanks,
Jen

 

[Doc Al]

What min strength is needed for a line that is to stop a fish of 85N in 0.112m if the fish is initially drifting at 2.5 m/s? (Assume constant deacceleration.)

What acceleration does this imply? And what force must the fish line exert to produce such an acceleration?

I'm having problems figuring out how to relate velocity to net x and y forces. I'm assuming tension is opposite of fish weight, but aside from that, I don't see any other forces??

I would assume that the vertical forces are in equilibrium so that all you need to worry about is the horizontal force due to the tension in the line. (Keep it simple.)

 

[j_confused]

Hrm. It might have helped that I didn't picture the fish and line similar to the classic elevator problem... (I don't fish.)

Acceleration, going back to vectors and 1-D motion, comes to -27.9 m/s^2; if I did things right. Does that seem too fast?

v(t) = Vo + at
0= 2.5 + at
t = -2.5/a

x(t) = Xo + Vot + 1/2at^2
plug in x(t) = 0.112, Xo = 0, t = -2.5/a
a = -27.9 m/s2

?

 

[Doc Al]

a = -27.9 m/s2

?

So far, so good.

 

[j_confused]

Alright, so now on to the net x-forces:
(W being weight, T being tension)

-W + T = ma

-85N + T = 85/9.8 x -27.9

then T = -156.99


... However, I didn't think tension could be negative, and 156N is the wrong answer.

Is W is the same direction of T then?

 

[Doc Al]

Alright, so now on to the net x-forces:
(W being weight, T being tension)

-W + T = ma

-85N + T = 85/9.8 x -27.9

then T = -156.99

The weight acts downward. But I would assume that the vertical forces are in equilibrium (the force of the water pushing the fish upward and the fish's weight pulling the fish down).

So I would say that the only unbalanced force on the fish is T = ma. (It's that simple.)

The sign (+ or -) of the acceleration or tension just refers to the direction that the acceleration or force acts.

 

[j_confused]

Wee ooo!

Thank you super mentor.

I guess I need to learn how to fish ^_^