# Moments question!

1. Jan 11, 2008

### hasnain721

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

2. Relevant equations

principle of moments

3. The attempt at a solution

I took moments about the base of teh retort stand :
(220 X x) = 228 X 9

x = 9.33 which is wrong as the question says show that it is 11 Newtons.

2. Jan 11, 2008

### Staff: Mentor

Where's the center of mass of the stand?

3. Jan 11, 2008

### hasnain721

in the middle i believe where the pivot it

4. Jan 11, 2008

### Staff: Mentor

I mean the center of mass of the retort stand alone. (Its center of mass cannot be over the pivot, otherwise it would tip over when the weight is added.)

5. Jan 11, 2008

### hasnain721

i thougt that the centre of mass is where the retort stand balances ..this is the full question :

6. Jan 11, 2008

### Staff: Mentor

Yes, in this diagram--given in part (b)--the center of mass is over the pivot point.

Now use that information when you find the moments in part (i).

7. Jan 11, 2008

### hasnain721

rite....moments is when u multiply the force by the perpendicular distance:

(220 X x) = (448-220) X 9

this does not work!

8. Jan 11, 2008

### Staff: Mentor

There are two forces that create torques on this thing. One is its weight. Where does that act? (Hint: Not at 220 mm from the pivot! Hint 2: Where's the center of mass of the retort stand?)

9. Jan 11, 2008

### hasnain721

the weight acts through the pivot and if we r taking moment about the pivot, i believe that it does not have a moment rite?

10. Jan 11, 2008

### Staff: Mentor

The weight of any object acts through its center of mass. Use the first diagram to find the center of mass of this object. (How far is the center of mass from the left end?)

11. Jan 11, 2008

### hasnain721

220 mm ?

12. Jan 11, 2008

### Staff: Mentor

I'm talking about the first diagram in your latest scan--the one marked part (b). 220 mm doesn't appear anywhere on that diagram.

13. Jan 11, 2008

### hasnain721

ah....40mm

14. Jan 11, 2008

### Staff: Mentor

Right. That diagram tells you where the center of mass of the object is. Now use that fact when you analyze the forces in part (i). Hint: How far is the center of mass from the pivot in that second diagram?

15. Jan 11, 2008

### hasnain721

srry i'm tryin my best...i'm rubbish at moments any ways :

if we take moments about the centre of mass : (40 X x) + ((448 - 40) X 9 ) = 0

stil doesnt work.

16. Jan 11, 2008

### Staff: Mentor

Take moments about the pivot point, not the center of mass. (Since the weight acts at the center of mass, it cannot exert a moment about the center of mass!)

17. Jan 11, 2008

### hasnain721

ok...if we take moments about the pivot then :

(40 X x) + (228 X 9 ) = 0 ?

18. Jan 11, 2008

### Staff: Mentor

Two problems:
(1) The distance between center of mass and pivot point is not 40. (That's the distance between center of mass and the left end.)
(2) Since one moment is clockwise and the other counterclockwise, they must have different signs.

19. Jan 11, 2008

### hasnain721

the centre of mass is at 40 mm and the pivot is 220 mm. Therefore the distance between them should be 220 - 40 = 180 mmm

yea clockwise moments = anticlock wise moments

(180 X x) = ((448 - 220) X 9 )

180 x = 228 X 9

180 x = 2052
x = 11.4 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

IT WORKS!!

THANK U SOOO MUCH!

20. Jan 11, 2008

### hasnain721

man u explain every thing so well.... ifu realy dun mind can u explain the next one plz :

i dont see how force and weight are related?

Last edited: Jan 11, 2008