Tension in legs of an insect dangling upside down

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Homework Help Overview

The problem involves an insect hanging from a horizontal rod, analyzing the tension in its legs while considering its mass and the angle at which it hangs. The discussion focuses on the relationship between the tension in the insect's legs and its weight, as well as how changes in leg position affect this tension.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the tension in the insect's legs and its weight, questioning the initial assumptions about the distribution of weight across the legs. Some suggest drawing free-body diagrams to clarify the forces at play.

Discussion Status

The discussion has progressed with some participants providing mathematical reasoning and exploring the implications of changing the angle of the legs. There is a recognition of the need for justification of claims made regarding tension and weight, and some productive guidance has been offered regarding the use of free-body diagrams.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available and the assumptions that can be made. The angle of the legs and its effect on tension is a key point of discussion.

coffeebird
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Homework Statement


Some insects can walk below a thin rod (such as a twig) by hanging from it. Suppose that such an insect has mass m and hangs from a horizontal rod as shown in Fig. 5-35, with angle θ = 40°. Its six legs are all under the same tension, and the leg sections nearest the body are horizontal. (a) What is the ratio of the tension in each tibia (forepart of a leg) to the insect's weight? (b) If the insect straightens out its legs somewhat, does the tension in each tibia increase, decrease, or stay the same?

Homework Equations



F=ma

The Attempt at a Solution

weight = 9.8 m = tension on one leg = 9.8/6 m

tension in all tibias = (9.8mcos40)^2 + (9.8sin40)^2

tension in all femurs= 9.8cos40=7.51m

in answer to (b), the tension in the tibias would decrease because the tension in the femurs would increase
 

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Last edited:
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Hello, coffeebird.

coffeebird said:
weight = 9.8 m = tension on one leg = 9.8/6 m

Did you really want to claim that the weight of the bug equals the tension in one leg? Or, are you claiming that the tension in one leg equals 1/6 the weight of the bug? Either way, you would need to justify that.

It might help to relate this problem to the more standard problem of finding the tension in two cords that support a hanging sign, as shown in the attached figure. The key to working problems like this is to draw a good free-body diagram of the sign. Likewise, for the bug, draw a free-body diagram of the bug treating the tibias as "cords" supporting the bug.
 

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(a) There are six legs, and the vertical component of the tension force in each leg is
T sinθ where θ = 40° . For vertical equilibrium (zero acceleration in the y direction) then by
Newton’s second law
6T sinθ= mg
∴ T = mg/6 sinθ
∴ the ratio becomes T/mg ≈ 0.259
(b) The angle θ is measured from horizontal, so as the insect “straightens out the legs” θ
will increase (getting closer to 90° ), which causes sinθ to increase (getting closer to 1)
and consequently (since sinθ is in the denominator) causes T to decrease.
 
thank you, i get it now
 

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