Center of mass with masses of unequal widths

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Homework Help Overview

The problem involves determining the center of mass of a barbell with weights of unequal distribution. The barbell is 1.5 m long, with three weights on one side and two on the other, each with a specified width and mass. The challenge lies in calculating the center of mass while considering the uniform mass of the bar itself.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the significance of the widths of the weights and their positions relative to the barbell's ends. There is an exploration of how to incorporate the mass of the bar into the calculations. Some participants question the method of measuring distances from the midpoint and the implications of unit errors.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning each other's reasoning. Some guidance has been provided regarding the inclusion of the bar's mass and the proper approach to measuring distances. There is no explicit consensus yet, but participants are engaging with the problem constructively.

Contextual Notes

Participants note the importance of correctly interpreting the problem's requirements, including the need to account for the bar's mass and the correct units of measurement. There is also mention of a provided answer from a study guide, which adds a layer of complexity to the discussion.

LBP
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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!
 
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Welcome to PF!

Hi LBP! Welcome to PF! :smile:
LBP said:
Then I subtracted 1.5m-.628m=.872

Why? :confused: …

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

(oh, and since the question asks for the distance from the midpoint, I'd have measured everything from there from the start)
 
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?
 
LBP said:
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. :confused:
… do I use its center of mass an weight as a M3X3?

yes! :smile:

(that's exactly what the centre of mass is defined for :wink:)
 
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.
 
LBP said:
… I know the answer is -5.62.

How can it be 5.62? :confused:

The barbell is only 1.5 long! :biggrin:​
 
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.
 
Your solution looks right to me. :confused:
 
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!
 

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