Moment about a point of a hammer

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SUMMARY

The discussion focuses on calculating the required force F to produce a clockwise moment of 590 lb⋅in about point A when pulling out a nail at point B using a hammer. The equation used is -590 = -Fcos(30)(18) - Fsin(30)(5), where the angle α between the force vector and the perpendicular to line segment XA must be determined geometrically. Participants emphasize the importance of using the correct angle and the efficiency of the method proposed by Robb for solving the problem.

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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?
 
So, F=72.9431?
 
Robb said:
So, F=72.9431?
Not what I get. Please post your working.
 
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
 
Robb said:
Sorry, -590=-F18cos(30)-5Fsin30)=32.6175
Yes.
 
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!
 
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haruspex said:
the method Robb used is valid, and quite efficient.
So it is. I hadn't noticed that.
 

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