What is the displacement of the beam after the medicine ball is thrown?

AI Thread Summary
The discussion centers on calculating the displacement of a beam when a medicine ball is thrown between two people standing at opposite ends. The center of mass for the system was correctly calculated as 1.65 m, which remains unchanged during the process. The initial calculation for the beam's displacement yielded x = 0.167 m, but further analysis revealed that the correct displacement is actually 0.175 m. This correction was confirmed through cross-checking the forces acting on the system. The importance of careful calculation and verification in physics problems is emphasized.
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Homework Statement


A person with mass m1 = 67.0 kg stands at the left end of a uniform beam with mass m2 = 91.0 kg and a length L = 3.3 m. Another person with mass m3 = 56.0 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 12.0 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.

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?)

I previously solved for the center of mass for the system, which came out to be about 1.65 m. This answer was correct, and I have been using it for my equations since.

Homework Equations



Xcm = [(m1)(x) + m2(x) + m3(x) + ... + mn(x)]/(m1 + m2 + m3 + ... + mn)

The Attempt at a Solution



I know that the center of mass does not change, so the system must act according to reorient itself. This means that the position of the beam is what changes. If we take the initial position of the person at the left end of the beam to be x = 0, this is the equation for the center of mass before the ball is thrown:

BEFORE:
Xcm = [(67)(0) + (91)(1.65) + (56)(3.3) + (12)(3.3)]/(67+91+56+12)
= 1.65 m

This is the center of mass. If we let x = the distance moved by the beam after the ball is thrown, we get the following equation:

AFTER:
1.65 = [(67)(x) + (91)(1.65+x) + (56)(3.3+x) + (12)(0.0 + x)]/(67+91+56+12)

And you simply solve for x to find the distance moved. The reason why the position of the medicine ball is now 0 in the after equation is because it was thrown to the left, and caught by the person on the left who was previously at the origin.

When I solve for x, I get x=0.167 m. I've double checked to make sure I'm not making any algebra mistakes or something like that. I feel like this should be the correct answer, but for some reason, it is wrong. Can anyone help me find out what I did wrong with my reasoning?
 
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WolframAlpha gives 1.657m for the center of mass and 0.175m for the displacement, using your formulas as input. Probably just a typo in the calculator?

Cross-check: (67+91+56)kg=214kg move x to the right, 12kg move 3.3m - x to the left. Solving 214*x=12*(3.3m-x) gives x=0.175m.
 
mfb said:
WolframAlpha gives 1.657m for the center of mass and 0.175m for the displacement, using your formulas as input. Probably just a typo in the calculator?

Cross-check: (67+91+56)kg=214kg move x to the right, 12kg move 3.3m - x to the left. Solving 214*x=12*(3.3m-x) gives x=0.175m.

That was right, thank you! I was worried I was making a mistake in my reasoning. I'm glad that wasn't it. I should double check my numbers before punching them into the calculator blindly next time. Thanks again! :D
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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