So the spider can safely climb up the strand with no acceleration (0 m/s^2).

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The discussion revolves around two physics problems involving forces and accelerations. The first problem involves a 2.0 kg fish that exerts a force of 40 N on a line rated for 38 N, leading to a calculated minimum acceleration of -1 m/s² to prevent the line from snapping. The second problem concerns a 0.00021 kg spider suspended from a web strand with a maximum tension of 0.002 N, resulting in a maximum safe climbing acceleration of 9.52 m/s², which contradicts the notion that the spider should remain stationary. The participants emphasize the importance of understanding net forces and tension in these scenarios.

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Help Exam Tmr! 112 Physics *FORCES*

As you may have guessed from the title, I'm in a bit of a bind here. You see the website with the answers to the review questions I'm working on isn't working... and there's an exam tomorrow. Here's a few of the questions any help would be greatly appreciated.
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A 2.0kg fish is hooked on a line that rated to a max of 38 N. At one point, the fish pulls with 40 N of force (fish are weightless in water). What is the min acceration at which you must play out the line to avoid it from snapping?
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A 0.00021 kg spider is suspended from a thin strand of webbing. The greatest tension the strand can withstand without breaking is 0.002 N . What is the max acceleration with which the spider can safetly climb up the strand?


answers
ok so apparently i need to show my work so here goes:

for the fish question:
m= 2.0kg
Fmax=38N
Fxrt= 40N
a = ?

Fnet canot be more than 38N soo..
Fnet=ma
38=2.0a
a=19m/s2

that doesn't seem right to me... awfully fast to make up for 2 N wouldn't you think?
orr wait..

Fnet needs to be -2? so then..

-2=2.0a
a= -1m/s2

WHAT IS GOING ON HERE

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See the problem with the spider question is this: the Fg of the spider ends up being the same as the max Force... but where Fg is also the net force i should be able to use F=ma... plug that in and you end up with 9.52m/s2... which seems to completely contradict that fact.?.?.?.
I mean wouldn't the spider only be able to hang there with no acceration at all, unless that is he goes down instead of up... so maybe the answer is -9.52?
 
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Force by the FISHING LINE (Tension) needs to be less than 38 N .
Force by the water (acting on the fish) is known.
Total Force acting on the fish (as your intuition had it) is 2N.
total F on fish = (m_fish)(a_fish)

The total Force acting on the spider looks very small,
maybe zero depending on the exact "g" value there.
so, T - mg = 0 = ma = 0