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 acceleration. For the fish question, the user grapples with calculating the minimum acceleration needed to prevent the fishing line from snapping, ultimately determining that the net force must not exceed 38 N. In the spider question, the user struggles with the relationship between tension and gravitational force, concluding that the spider can only safely climb with zero acceleration, as any upward movement would exceed the strand's tension limit. The confusion stems from the calculations leading to seemingly contradictory acceleration values. The key takeaway is that both scenarios highlight the importance of understanding net forces and tension limits in physics problems.
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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
 

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