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

AI Thread Summary
The discussion revolves around calculating the total work required to drive a screw into wood, with the problem stating that it takes 25 turns and the maximum torque is 15 N m. The friction force increases linearly with the screw's depth, necessitating the use of integration to find the work done. Participants clarify that the torque is not constant, leading to confusion about the angle of rotation in relation to the number of turns. The correct approach involves defining the torque as a function of depth and translating it into the angle of rotation. Ultimately, the revised equation for work integrates the torque function over the correct range of angles.
Gingrbreadman1
Messages
3
Reaction score
0

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
 
Physics news on Phys.org
Welcome to PF,

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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top