• December 28th 2006, 09:05 AM
anthmoo
I was on this question for over 45 minutes and I was stumped! I really have no idea what to do! Please help!

Question

http://anthmoo.fromtheabyss.com/damnhelp.JPG

The diagram above shows a pillar (lying horizontally) made of two uniform sections X and Y each of cross-sectional area $3.5*10^-2 m^2$. The sections are made from two different materials. The weights of X and Y are shown in the diagram acting through the centre of gravity of each section.

The pillar will balance horizontally when supported vertically below point P.

Show, using the principle of moments, the the point P is 1.2m from the end of the pillar (the right side of section Y).

Thanks guys!
• December 28th 2006, 11:18 AM
ThePerfectHacker
Quote:

Originally Posted by anthmoo
I was on this question for over 45 minutes and I was stumped! I really have no idea what to do! Please help!

Question

http://anthmoo.fromtheabyss.com/damnhelp.JPG

The diagram above shows a pillar (lying horizontally) made of two uniform sections X and Y each of cross-sectional area $3.5*10^-2 m^2$. The sections are made from two different materials. The weights of X and Y are shown in the diagram acting through the centre of gravity of each section.

The pillar will balance horizontally when supported vertically below point P.

Show, using the principle of moments, the the point P is 1.2m from the end of the pillar (the right side of section Y).

Thanks guys!

Something I do not understand.
How can each have the same area?
Perhaps, you mean the large one?
• December 28th 2006, 11:32 AM
CaptainBlack
Quote:

Originally Posted by anthmoo
I was on this question for over 45 minutes and I was stumped! I really have no idea what to do! Please help!

Question

http://anthmoo.fromtheabyss.com/damnhelp.JPG

The diagram above shows a pillar (lying horizontally) made of two uniform sections X and Y each of cross-sectional area $3.5*10^-2 m^2$. The sections are made from two different materials. The weights of X and Y are shown in the diagram acting through the centre of gravity of each section.

The pillar will balance horizontally when supported vertically below point P.

Show, using the principle of moments, the the point P is 1.2m from the end of the pillar (the right side of section Y).

Thanks guys!

The moments of the weights about P will sum to zero if the pillar balances at P (algebraic sum where the sign of one is poitive and the other negative).

Take the right end of the pillar to be x=0 m, then the centre of mass of the
first segmant is at x=0.4 m, and the second is at x=1.4 m. Thus for balance

250(P-0.4)=1000(1.6-P)

and I make the solution of this P=1.12 m.
• December 28th 2006, 01:05 PM
anthmoo
Quote:

Originally Posted by ThePerfectHacker
Something I do not understand.
How can each have the same area?
Perhaps, you mean the large one?

I think you should ignore that, it was for a previous part of the question. :D
• December 28th 2006, 01:06 PM
anthmoo
Quote:

Originally Posted by CaptainBlack
The moments of the weights about P will sum to zero if the pillar balances at P (algebraic sum where the sign of one is poitive and the other negative).

Take the right end of the pillar to be x=0 m, then the centre of mass of the
first segmant is at x=0.4 m, and the second is at x=1.4 m. Thus for balance

250(P-0.4)=1000(1.6-P)

and I make the solution of this P=1.12 m.

Apparently that is wrong :( it is 1.2m as it is said in the question...

:confused:
• December 28th 2006, 02:10 PM
CaptainBlack
Quote:

Originally Posted by anthmoo
Apparently that is wrong :( it is 1.2m as it is said in the question...

:confused:

That could be because of a mistake in my arithmetic in calculating the position for the 1000 N
weight It should be 1.2/2+0.8=1.4 m. I just got the sum wrong (that is I had 1.6m) It now looks
like it should be:

250(P-0.4)=1000(1.4-P)

which has solution P=1.2

RonL
• December 28th 2006, 04:57 PM
anthmoo
Quote:

Originally Posted by CaptainBlack
That could be because of a mistake in my arithmetic in calculating the position for the 1000 N
weight It should be 1.2/2+0.8=1.4 m. I just got the sum wrong (that is I had 1.6m) It now looks
like it should be:

250(P-0.4)=1000(1.4-P)

which has solution P=1.2

RonL

Thankyou very much CB! You are a life saver yet again :D
• December 29th 2006, 03:02 PM
ticbol
Quote:

Originally Posted by anthmoo
I was on this question for over 45 minutes and I was stumped! I really have no idea what to do! Please help!

Question

http://anthmoo.fromtheabyss.com/damnhelp.JPG

The diagram above shows a pillar (lying horizontally) made of two uniform sections X and Y each of cross-sectional area $3.5*10^-2 m^2$. The sections are made from two different materials. The weights of X and Y are shown in the diagram acting through the centre of gravity of each section.

The pillar will balance horizontally when supported vertically below point P.

Show, using the principle of moments, the the point P is 1.2m from the end of the pillar (the right side of section Y).

Thanks guys!

Principle of moments?
Is that, "the moment of a force about a point is the product of the force and its perpendicular distance from the point"?
Or is that, "in a system in static equilibrium, the summation of moments about a point is zero"? Or, clockwise moments equal counterclockwise moments, so no movement?

Whatever.
Here is one way of solving your problem here.

In the figure, if the beam will balance horizontally if supported vertically below point P, then P is vertically in line with the centroid of the beam. So, the total weight of the beam can be imagined as concentrated at, or vertically with, point P.
So, the external force, F, to balance the beam vertically below point P is
F = Resultant of downward forces = 1000 +250 = 1250 N.

Summation of moments about the lower righthand corner of the composite figure is zero.
Let x = horizontal distance of point P, and so of force F=1250N, from the lower righthand corner, in meter.
Clockwise moment is negative; counterclockwise moment is positive.
[250(0.8 /2)] -[1250(x)] +[1000(0.8 +(1.2 /2))] = 0
100 -1250x +1400 = 0
x = -1500 / (-1250)