Moment about a point of a hammer

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Homework Statement



Probs._48_9.jpg

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.

Homework Equations

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.
 
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Robb said:
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.
How would you solve 3=x+2x?
 
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.
 
haruspex said:
Not what I get. Please post your working.
Sorry, -590=-F18cos(30)-5Fsin30)=32.6175
 
andrewkirk said:
The angle you need to use is not 30 degrees but the angle αα\alpha between the force vector and the perpendicular to the line segment ¯¯¯¯¯¯¯¯¯XAXA¯\overline{XA}. You can calculate that angle using geometry, from the information given.
the method Robb used is valid, and quite efficient.
 
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haruspex said:
Yes.
Dang, nothing like making a mountain out of a mole hill!