Solve Moment Problem: Weight of Bar = 350 N

[urbanupstart]

1. A heavy uniform bar is in equilibrium in the arrrangement shown
(I cannot upload the picture, but the bar is resting on a pivot with two weights at either end. 1.5m to the left of the pivot with 1000N weight at the end. 4.5m to the right of the pivot with a 100N weight at the end)
Question is- what is the weight of the bar to nearest Newton?

Homework Equations

3. My attempt so far- anti-clockwise moment of weight is 1.5*1000 = 1500Nm; clockwise moment of right weight is 100* 4.5 =450Nm.

I have never met a question asking me about the bar, but... I reasoned that as the bar is balanced, the clockwise and antclockwise moments must be the same and so
1500Nm + 1.5x = 450Nm + 4.5x (the 1.5x and 4.5x refer to the bar moments/weights)

simplifying to 1050Nm = 3.5x
so x = 350

I have the answer- the bar is 300N,so I was close:) but I cannot get it. Anyone help me through this please?

 

[tiny-tim]

welcome to pf!

hi urbanupstart! welcome to pf!

1500Nm + 1.5x = 450Nm + 4.5x (the 1.5x and 4.5x refer to the bar moments/weights)

what is this?

there are three forces:

the weights of the two weights, and the weight of the bar itself …

add the moments of those three forces ​

 

[urbanupstart]

what is this?

there are three forces:

the weights of the two weights, and the weight of the bar itself …

add the moments of those three forces ​

Ah, that was my attempt to figure out the bar weight lol!

Thanks Tim, My level of understanding of this is pretty low though. How would I calculate the bar moment? Would you be kind enough to show me the calcs, so I can take a look please. Sorry to be so dense

 

[tiny-tim]

hi urbanupstart!

(just got up :zzz: …)

all you have to do is assume that the weight of the bar is a force which goes through its centre of mass

 

[urbanupstart]

I'm so sorry, but I mis-read the answer. It should be 700N Can anyone still help me with this please?

 

[tiny-tim]

ok, assume that the weight of the bar is is w, and that it goes through the centre of mass of the bar …

write out the moments of all three weights ​

 

[urbanupstart]

Ok, that would be 1500Nm(for the first), I don't know how to figure out the bar moment (as distance is zero and zero* anything is zero), but on the other hand, if it is the sum of the other two moments, then it would be 1950Nm? and 450Nm(for the right-hand weight)

 

[tiny-tim]

Ok, that would be 1500Nm(for the first), I don't know how to figure out the bar moment (as distance is zero and zero* anything is zero), but on the other hand, if it is the sum of the other two moments, then it would be 1950Nm? and 450Nm(for the right-hand weight)

if you're taking moments about the pivot, you need to multiply the weight of the bar by the distance (which isn't zero) from the pivot to the centre of mass of the bar

 

[urbanupstart]

ok, then the moment of the bar is w*6m?

 

[tiny-tim]

ok, then the moment of the bar is w*6m?

no, 6 m is the whole length of the bar, the moment about the pivot only uses the distance from the c.o.m. to the pivot

 

[urbanupstart]

ok, then w*4.5 + w* 1.5 ?

 

[tiny-tim]

ok, then w*4.5 + w* 1.5 ?

uhh? how far is the c.o.m. of the bar from the pivot?

 

[urbanupstart]

Hmmm, centre of mass of the bar from the pivot? I don't know, please help

 

[tiny-tim]

… the bar is resting on a pivot with two weights at either end. 1.5m to the left of the pivot with 1000N weight at the end. 4.5m to the right of the pivot with a 100N weight at the end …

Hmmm, centre of mass of the bar from the pivot? I don't know, please help

the bar is 6 m long, so the c.o.m. is 3 m from either end, and the pivot is 1.5 m from one end

and now I'm going to bed :zzz:

 

[urbanupstart]

Ok, I appreciate your reponses Tinytim, but I still need more help.

Please go on...

 

[tiny-tim]

(just got up :zzz: …)

from the pivot to one weight is +4.5 m, to the c.o.m. is +1.5 m, and to the other weight is -1.5m

those are the distances you should be using in your moments