# Wrench and nut

1. Mar 5, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
If a force a F acts on the farther end of the wrench, the torque due to it Fl where l is the length of wrench. The work done by this torque for one full turn is Fl*(2$\pi$)
For wrench A, l=10 cm, and for wrench B, l=20cm.
Therefore the ratio of work done is 0.5. But this is wrong.
I don't understand why they have given the radius of the nut. I haven't used this information and I have no idea about where to use this.

Any help is appreciated. Thanks!

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2. Mar 5, 2013

### Staff: Mentor

If a force F is required to tighten the nut using the 10 cm wrench, what force is required with the 20 cm one?

3. Mar 5, 2013

F/2?

4. Mar 5, 2013

Right.

5. Mar 5, 2013

### Saitama

But what next? Does that mean the ratio is 1?

6. Mar 5, 2013

### Staff: Mentor

That's what I would say.

7. Mar 5, 2013

### Saitama

Thanks a lot Doc Al!