Verify Moment Calculation: Attached PDF Answer

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Homework Help Overview

The discussion revolves around the calculation of moments in a physics problem involving forces and their respective distances from a pivot point. Participants are examining the application of the moment equation, M = Fd, in the context of a provided PDF file.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to verify moment calculations for different forces acting on a pivot point. Questions arise regarding the signs of moments and the correct application of the moment equation. There is also discussion about the components of forces and their contributions to the overall moment.

Discussion Status

The discussion is active, with participants providing insights and corrections regarding the calculation of moments. Some guidance has been offered on the importance of considering the perpendicular distances and the signs associated with clockwise and counterclockwise moments. Multiple interpretations of the force components and their effects on the moment are being explored.

Contextual Notes

Participants are working with a specific problem setup that includes forces and distances, but there are indications of potential confusion regarding the orientation of force components and their respective contributions to the moment about the pivot point. The original poster has provided a PDF for reference, which is central to the discussion.

drkidd22
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Homework Statement



attached .pdf file

Homework Equations



M = Fd

The Attempt at a Solution



See attached pdf please

I would like for someone to verify my answer, thanks
 

Attachments

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drkidd22 said:

Homework Statement



attached .pdf file

Homework Equations



M = Fd

The Attempt at a Solution



See attached pdf please

I would like for someone to verify my answer, thanks
You are not determining moments and the signs of the moments correctly. The moment of a force (or force component) is the the product of that force times the perpendicular distance from its line of action to the pivot point. Watch plus and minus signs.
 
So let's take F1. Wouldn't F1 be 10lb * 8in = +80 lb*in CCW?
 
drkidd22 said:
So let's take F1. Wouldn't F1 be 10lb * 8in = +80 lb*in CCW?
Yes, the moment from F1 about A is 80 in-lb ccw. Now move on to F2 and F3.
 
F2x = 10Sin60 = 8.66lb-in cw (which will be negative)
F2y = 10Cos60 = 5lb-in cw (which will be negative)

F3x = 10lb * 10in = 100lb-in ccw
 
drkidd22 said:
F2x = 10Sin60 = 8.66lb-in cw (which will be negative)
F2y = 10Cos60 = 5lb-in cw (which will be negative)

F3x = 10lb * 10in = 100lb-in ccw
Don't confuse force with moment. F3 is 10 lbs, so the moment of F3 about A is your correct answer of 100 in-lb.

Now for F2, you broke it up into its x and y components (although you may have the x and y-axis reversed).. Bit you failed to calculate the moments from each of those component forces, using M = force times perpendicular distance.

(The unit of moment in the USA is , historically, commonly called inch-pound, not pound-inch, although it's the same. )
 
I see what you mean for F2.

What I really solved for was:
F2x = 8.66lb
F2y = 5lb

but I don't understand about having the components reversed
 
drkidd22 said:
I see what you mean for F2.

What I really solved for was:
F2x = 8.66lb
F2y = 5lb

but I don't understand about having the components reversed
Sorry, you had it correct, using the x-axis as the horiziontal axis, F2x is F2 cos 30, or F2 sin 60, same thing, and F2y is F2 sin 30, or F2 cos 60, same thing. Now what is the moment about A from F2x? What is the moment about A from F2y?
 
F2x = 8.66lb * 12in = 103.92 in-lb
F2y = 5.00lb * 12in = 60.000 in-lb
 
  • #10
drkidd22 said:
F2x = 8.66lb * 12in = 103.92 in-lb
F2y = 5.00lb * 12in = 60.000 in-lb
Let's start with F2y. The moment of F2y about A is 5(12) = 60 in-lb. Clockwise or ccw? Plus or minus?
Now your calculation for the moment of F2x about A is incorrect. Remember, M = force times perpendicular distance. So what's the pependicular distance from the line of action of F2x to point A?? It is not 12 inches.
 
  • #11
F2y would be Clockwise.
F2x is less than 12 inches, but I'm not sure how to solve for that or there is simply no moment as F2x can't cause moment, it won't be perpendicular?
 
Last edited:

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