# Moment about a point of a hammer

1. Oct 6, 2016

### Robb

1. The problem statement, all variables and given/known data

In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of 590 lb⋅in about point A.(Figure 1)

Determine the required magnitude of force F.
2. Relevant equations

3. The attempt at a solution

M=590lb*in

-590= -Fcos(30)(18)-Fsin(30)(5)

I'm not sure how to deal with F. I know it's simple if given the force to determine the moment but I'm a bit stumped trying to go the other way.

2. Oct 6, 2016

### haruspex

How would you solve 3=x+2x?

3. Oct 6, 2016

### Robb

So, F=72.9431?

4. Oct 6, 2016

5. Oct 6, 2016

### andrewkirk

Let $X$ be the point where the force is applied to the handle.

The angle you need to use is not 30 degrees but the angle $\alpha$ between the force vector and the perpendicular to the line segment $\overline{XA}$. You can calculate that angle using geometry, from the information given.

The Torque (moment) will be the force multiplied by $\cos\alpha$ and the length of $\overline{XA}$.

By the way, that's one weird hammer! Based on the drawing and the dimensions given, the claw is about eight inches long! That's a gemmy, not a hammer. The drawing is not to scale. The 18 in length is not 18/5 times the 5 inch length. That's why the hammer doesn't look weird.

6. Oct 6, 2016

### Robb

Sorry, -590=-F18cos(30)-5Fsin30)=32.6175

7. Oct 6, 2016

Yes.

8. Oct 6, 2016

### haruspex

the method Robb used is valid, and quite efficient.

9. Oct 6, 2016

### Robb

Dang, nothing like making a mountain out of a mole hill!

10. Oct 6, 2016

### andrewkirk

So it is. I hadn't noticed that.