# Homework Help: Moment calculation

1. May 5, 2010

### drkidd22

1. The problem statement, all variables and given/known data

attached .pdf file

2. Relevant equations

M = Fd

3. The attempt at a solution

I would like for someone to verify my answer, thanks

File size:
319.4 KB
Views:
142
2. May 5, 2010

### PhanthomJay

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.

3. May 5, 2010

### drkidd22

So lets take F1. Wouldn't F1 be 10lb * 8in = +80 lb*in CCW?

4. May 5, 2010

### PhanthomJay

Yes, the moment from F1 about A is 80 in-lb ccw. Now move on to F2 and F3.

5. May 5, 2010

### drkidd22

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

6. May 5, 2010

### PhanthomJay

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. )

7. May 5, 2010

### drkidd22

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

8. May 5, 2010

### PhanthomJay

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?

9. May 5, 2010

### drkidd22

F2x = 8.66lb * 12in = 103.92 in-lb
F2y = 5.00lb * 12in = 60.000 in-lb

10. May 5, 2010

### PhanthomJay

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. May 5, 2010

### drkidd22

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: May 5, 2010