How Does Throwing a Medicine Ball Affect Beam Displacement?

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Homework Help Overview

The discussion revolves around a physics problem involving a uniform beam and the effect of throwing a medicine ball on the beam's displacement. The problem includes specific masses and positions of individuals on the beam and requires determining the new position of a person after the ball is thrown.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the center of mass and its implications for the beam's displacement. There are attempts to verify the center of mass value, with some participants questioning the accuracy of their calculations and the accepted values.

Discussion Status

The discussion is ongoing, with participants sharing different center of mass values and their experiences with the online system's acceptance of answers. There is no explicit consensus on the correct value, but some guidance has been offered regarding the calculations.

Contextual Notes

Participants are working within the constraints of an online homework system that may have specific answer requirements, leading to confusion about the correctness of their calculations.

tgitatif
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1. Homework Statement
A person with mass m1 = 61 kg stands at the left end of a uniform beam with mass m2 = 104 kg and a length L = 2.7 m. Another person with mass m3 = 68 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 8 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

3)After the ball is thrown to the person on the left, what is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?)


2. Homework Equations
Center of mass equation (can't figure out how to type it on here)

3. Attempt at a solution
I kept the original center of mass of 1.53 since it doesn't move and created an equation to solve for the new position.

1.53 = \frac{(61+8) (0-x) + 104 (1.35-x) + 68 (2.7-x)}{241}

And I get x = .186m, which is not correct.
 
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I would check your center of mass value to start -- I got 1.434.
 
The system accepted 1.53 as the answer. I could try 1.434 though.
 
Wow. I don't know why it accepted something so far off for the first question. Using 1.43 worked, thanks.
 
No problem -- those darn online things.
 

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