How Much Work Does It Take to Drive a Screw into Wood?

[smhippe]

Homework Statement

It takes 20 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 12 N m
when the screw is completely in the wood, what is the total work (in Joules) required to
drive in the screw?

The Attempt at a Solution

So I tried to do an integral by finding out how much the torque changed per rotation. Then using that as the equation and the total distance turned to be plugged in.
[tex]\int.6x[/tex] from 0 to 7200. I got a very large number and I don't think I did it right...

 

[anathema]

Hello,

Thank you for your post. I would like to offer some clarification and assistance with your attempt at a solution.

Firstly, your integral seems to be incorrect. The torque required for turning the screw increases linearly with the depth the screw has penetrated, so the equation should be T = kx, where k is the constant of proportionality and x is the depth of penetration. This means that the torque required for each rotation will be different, as the depth of penetration increases with each turn.

To find the total work required to drive in the screw, we need to integrate the torque equation with respect to x, from 0 to the maximum depth of penetration (in this case, 20 turns or 7200 degrees). This will give us the area under the torque curve, which represents the total work done.

However, we need to know the value of k in order to solve the integral. To find this, we can use the given information that the maximum torque is 12 N m when the screw is completely in the wood. This means that when x = 20 turns, T = 12 N m. Plugging these values into our equation, we get 12 = k(20), which gives us k = 0.6 N m/turn.

Now, we can integrate the torque equation \int 0.6x dx from 0 to 7200 to get the total work required. This gives us a value of 2592 Joules.

I hope this helps clarify the solution for you. Please let me know if you have any further questions or concerns. Keep up the good work!