Center of mass with masses of unequal widths

[LBP]

Homework Statement

A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left, and two weights of the same mass are hung on the right. The width of each weight is 4 cm and the outside edge of each group of weights is placed 4 cm from the ends. Where is the center of mass of the barbell as measured from the mid-point, M, of the bar? The bar is of uniform mass and has mass 5 kg, and the retaining collars are of negligible mass. Take to the right as positive.

Homework Equations

Center of mass= (m1x1+m2x2)/(m1+m2)

The Attempt at a Solution

So I know that the widths of the masses on each side are likely to play a role here. Knowing that the masses on the left total a width of 12 cm, the center of mass of just the weights would be 6 cm. Then since the problem states they are positioned 4 cm inward I added 4 cm +6 cm= 10 cm or .1M and used that as X1. Then on the right the width is 8 cm so the center is 4 cm moved over by 4 cm so the position is 8 cm left of the right most end. To find X2 I took 1.5 m - .08 M =1.42 M. Then I took the masses and the weights and put them in the formula

((.1m)(60kg)+(1.42)(40.kg))/(60kg+40kg)=.628 M

Then I subtracted 1.5m-.628m=.872
The half way point on the bar is .75 so the center of gravity would be .122 to the right of the half way point.
I realize this doesn't make sense because the center point should be closer to the heavier side. I'm just not sure what I did wrong.

Please help! Thanks!

 

[tiny-tim]

Welcome to PF!

Hi LBP! Welcome to PF!

Then I subtracted 1.5m-.628m=.872

Why? …

you were doing fine until then (except that you left out the weight of the bar ).

(oh, and since the question asks for the distance from the midpoint, I'd have measured everything from there from the start)

 

[LBP]

I was thinking subtracting from 1.5 would give me the position on the bar and from that figure out that position's distance from the center, no?

Where does the weight of the bar go? Do I add the weight to either position? or do I use its center of mass an weight as a M3X3?

 

[tiny-tim]

I was thinking subtracting from 1.5 would give me the position on the bar and from that figure out that position's distance from the center, no?

Sorry, not followoing you.

… do I use its center of mass an weight as a M3X3?

yes!

(that's exactly what the centre of mass is defined for )

 

[LBP]

humm Okay so I reworked the problem. Knowing the left mass is .1M in from the left and the right mass is .08 M in from the right and the middle is .75M. Then X1= -.65 X2=0 and X3=.67

So I did ((-.65*60)+(0*5)+(.67*40))/(40*60+5)= -.116 but I know the answer is -5.62.

 

[tiny-tim]

… I know the answer is -5.62.

How can it be 5.62?

The barbell is only 1.5 long! ​

 

[LBP]

Oh I am getting careless with units! -5.62 cm not meters sorry! The question is from a study guide for which we were given the answers.

 

[tiny-tim]

Your solution looks right to me.

 

[LBP]

I was afraid of that haha. I appreciate your help anyhow. I have a better understanding of what's going on then when I started, so thank you!