Solve Min Acceleration for 5g Spider: Gr 11 Physics

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The discussion centers on calculating the minimal acceleration for a 5g spider lowering itself on a web with a tensile strength of 0.03N. The spider's weight generates a gravitational force of 0.05N, which exceeds the web's capacity if the spider remains still, causing the web to snap. To avoid this, the spider must accelerate downward to reduce the tension on the web below its breaking point. The net force acting on the spider is determined to be 0.02N, leading to a calculated minimal acceleration of approximately 4m/s² to prevent the web from breaking. The conversation emphasizes the importance of understanding forces and drawing diagrams to visualize the problem.
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Hi everyone, I have another question which troubles me. It says
" A 5g spider is lowering itself to the floor. Its web has a tensile strength of 0.03N. What is the minimal acceleration with which the spider can lower itself?"

Well my question is what do they mean by minimal acceleration and what equation can I use to find it. Just to let you guys know I'm only in Gr. 11 physics so I won't understand any complicated equations that are used in Universities or higher grades. Thanks.
 
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No complicated equations are needed, this is more like a conceptual thing.

Draw a diagram.
Why can't the spider stay still on the web? What is N, the force exerted by the web, if the spider does not move?
 
Well what I'm trying to find is the minimal acceleration of the spider so I think the spider have to move to have acceleration?
 
True, that's the info given in the problem. If the spider would, however, stay still, what would happen? Why?
 
could you give me a little more hint at what your getting at?
 
The web would snap, as it can only hold 0,03 N and the spider weighs 0,005 kg * 9,81 m/s^2.

If you already haven't, draw a diagram with all the forces that are acting on the spider.
 
so what you are saying is that the force of gravity acting on the spider is 0.05N which is 0.02 N more then the web's capacity which in term causing it to snap?

So is this a trick question in which I can't find an actual minimal acceleration?
 
PhysicBeginner said:
so what you are saying is that the force of gravity acting on the spider is 0.05N which is 0.02 N more then the web's capacity which in term causing it to snap?
Yes.
PhysicBeginner said:
So is this a trick question in which I can't find an actual minimal acceleration?
No.


Please, draw a diagram. The forces acting on the spider are N (which I explained earlier) and G, the force caused by gravity (plus air resistance, but that doesn't count). We know that F (the net force) = ma (Newton II). What is the net force?
 
PhysicBeginner said:
So is this a trick question in which I can't find an actual minimal acceleration?

No. The web will snap if the spider sets still, but the Spider is accelerating downwards and this will reduce the tension on the web. (example: If the Spder were to accelerate down the thread at 9.8m/sec2, then this would be the same as if he were freefalling, and he would put zero tension on the thread.)

The question is asking what the least downward acceleration he must have in order to reduce the tension on the thread to below its breaking point.
 
  • #10
so the acceleration would be smaller than 9.8m/s^2 right?
So I just have to find the acceleration which gives a Force smaller than that of the thread so it wouldn't snap?

like
0.29=0.0005*a
a=5.8m/s^2 ?
 
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  • #11
PhysicBeginner said:
like
0.29=0.0005*a
a=5.8m/s^2 ?
No. If you set a = g, tension should be 0.

As I've said earlier: draw a diagram.
Can you figure out the net force acting on the spider?
 
  • #12
The net force is 0.02N[Down]=0.05N[Down] (The Force of gravity acting on the spider)-0.03N[up] (The Force of the tread pulling the spider up) ?
 
  • #13
Yep, that looks more like it.

(For greater accuracy: Instead of using 0,05 N, use 0,005 kg * 9,81 m/s^2).
 
  • #14
So now that I've found the net force acting on the spider do I then find the acceleration with the net force like this?

0.02N=0.005*a
a=4m/s^2
so the minimal acceleration is 4m/s^2 which eliminates the extra 0.02N acting on the thread?
 
  • #15
Correct. (Actually 3,81 m/s^2, if you use 9,81 instead of 10 for gravity, but that rounds to 4 m/s^2.)
 
  • #16
Yes I got it at last. Thanks Paallikko for all your patience and time. Thank you soooooooo much!
 
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