Moment about a point of a hammer

  • Thread starter Robb
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  • #1
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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.
 

Answers and Replies

  • #2
haruspex
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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?
 
  • #3
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So, F=72.9431?
 
  • #5
andrewkirk
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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
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Not what I get. Please post your working.


Sorry, -590=-F18cos(30)-5Fsin30)=32.6175
 
  • #8
haruspex
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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.
 
  • #10
andrewkirk
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the method Robb used is valid, and quite efficient.
So it is. I hadn't noticed that.
 

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