What is the Total Work Required to Drive a Screw Completely into Wood?

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Homework Help Overview

The problem involves calculating the total work required to drive a screw into wood, considering the relationship between torque, depth of penetration, and friction. The context is rooted in mechanics, specifically focusing on rotational motion and work done against friction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the number of turns and the angle of rotation, questioning the initial assumption that 25 turns corresponds directly to 25 radians. There is an exploration of the torque function and its dependence on depth, with suggestions to derive the torque as a function of depth and then relate it to the angle.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the torque function and its variability with depth. Some have attempted to adjust their equations based on feedback, indicating a productive exploration of the problem's complexities.

Contextual Notes

There is a noted confusion regarding the relationship between the number of turns and the corresponding angle in radians, as well as the need to correctly interpret the torque function in relation to the depth of the screw's penetration.

Gingrbreadman1
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Homework Statement



It takes 25 turns to drive a screw completely into a block of wood. Because the friction force between the wood and the screw is proportional to the contact area between the wood and the screw, the torque required for turning the screw increases linearly with the depth that the screw has penetrated into the wood. If the maximum torque is 15 N m when the screw is completely in the wood, what is the total work (in J) required to drive in the screw?

Homework Equations



Work = ∫torque dθ

The Attempt at a Solution



Work = ∫(15/25)x dθ (from 0 to 25)
Work = 187.5
 
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Uh, 25 turns does not correspond to theta = 25. Think about it. What angle do you go through in one turn?
 
One revolution would be 2PI, but when I do the same equation with the full 50PI I still get the wrong answer.
 
Gingrbreadman1 said:
One revolution would be 2PI, but when I do the same equation with the full 50PI I still get the wrong answer.

Well that's no surprise. You've got your integrand being constant. The whole point of this question is that that the torque ISN'T constant with angle. If it was constant, it wouldn't be necessary to integrate at all!

You have to think more carefully about what the function τ(θ) is. If I were you, I'd find the function τ(x), where x is the depth into the wood, and then translate this into τ(θ).
 
Thanks, changed the equation to the integral of (15/(2PI * 25))x from 0 to 50PI
 
Gingrbreadman1 said:
Thanks, changed the equation to the integral of (15/(2PI * 25))x from 0 to 50PI

I think that's correct, provided x is actually meant to be θ in that equation.
 

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