How Fast Can a Spider Lower Itself Without Breaking Its Web?

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A 5.00 g spider is lowering itself on a web with a tensile strength of 3.00 x 10^-2 N. To determine the minimal acceleration for the spider without breaking the web, Newton's second law is applied, focusing on the tension in the web. The tension must equal the gravitational force acting on the spider, which allows for the calculation of acceleration. The discussion clarifies that there is no normal force involved in this scenario, simplifying the analysis to just the gravitational force and the web's tensile strength. The critical tension value is essential for solving the problem accurately.
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a 5.00 g spider is lowering itself to the floor. Its web has a tensile strength of 3.00 x 10-2N. What is the minimal accerleration with which the spider can lower itself?


So i don't know how to begin, would the tensile strength of the web be equal to the total forces? as it is equal to force of gravity and normal force? what would the steps be and how to get the answer and what's the answer? thanks a bunch! :smile:
 
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Hi bobsagget! :smile:

Use good ol' Newton's second law to find how the tension in the web depends on the acceleration. :wink:
 
so would the force of gravity and the normal force need to equal the total tensile strength of the web? then i can solve for acceleration after.
 
bobsagget said:
so would the force of gravity and the normal force need to equal the total tensile strength of the web? then i can solve for acceleration after.

What normal force? :confused:

Apply Newton's second law !
 
ok so no normal force, so how would i draw that as a fbd though? the forces are 1) force of gravity down(mg), and anything else? or would there just be on and then i have to find the acceleration if the total force applied has to be equal to that of the webs strength
 
bobsagget said:
… and anything else? …

Yes, the tension!
 
and that tension is the 3.00 x10-2
 
(just got up :zzz: …)

Yes, that's the critical tension (just before the web breaks), so that's the value you should use in your calculations.

What do you get? :smile:
 
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