Moment about a point of a hammer

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Homework Help Overview

The problem involves calculating the force required to produce a specific moment about a point using a hammer. The context includes the application of torque and the geometry of the hammer's handle and angle of force application.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between force and moment, with some expressing confusion about rearranging the equation to solve for force. There are questions about the correct angle to use in the calculations and the geometry involved.

Discussion Status

Some participants have suggested alternative angles and methods for calculating the moment, while others have expressed uncertainty about the calculations presented. There is an ongoing exploration of the problem without a clear consensus on the correct approach.

Contextual Notes

Participants note discrepancies in the drawing and dimensions provided, suggesting that the representation may not accurately reflect the physical scenario. There is also mention of homework constraints that may limit the information available for solving the problem.

Robb
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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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