Simple vector train problem drawing

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In the discussion about a ball hanging from a string in a train car moving at a constant velocity, the key focus is on creating a correct free-body diagram. The user initially drew two equal-length vectors that are 180 degrees apart, representing forces acting on the ball, but received feedback indicating the diagram was incorrect. The user speculates that the inclusion of friction forces between the train and tracks might be necessary for a complete analysis. They also consider whether the weight vector should be labeled as 'mg' instead of 'w'. The conversation highlights the importance of accurately representing all forces in a free-body diagram for systems in equilibrium.
kirby27
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A ball is hanging from a long string that is tied to the ceiling of a train car traveling eastward on horizontal tracks. An observer inside the train car sees the ball hang motionless.

Draw a clearly labeled free-body diagram for the ball if the train has a uniform velocity.


i drew two vectors, but it says i got it wrong. both of my vectors are equal in lenth and are 180 degrees apart.

here's my answer image: http://i53.tinypic.com/21et2za.jpg
 
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Maybe my intro mechanics are a bit rusty, but I can't see why that would be incorrect. If the system is in equilibrium (Fnet = 0) that means that the train is moving at a constant velocity (i.e., anet = 0). Perhaps since there is a friction force between the tracks and the train, and (by virtue of Fnet being 0) the train exerts a forward force to offset it, they want you to add vectors Ftrain = \muFgi and Fk = \muFg(-i)

where i is the unit vector in the direction of the train's velocity vector.

That's the best I can come up with.
 
Looks good, maybe they want 'mg' instead of 'w' for the weight vector?? I don't know why it doesn't accept it, though.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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