Solving the Mystery of Standing Still in a Moving Train

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Homework Help Overview

The discussion revolves around a physics problem involving a passenger standing in a forward accelerating train. The problem requires drawing free body diagrams (FBDs) from two different frames of reference: the Earth's frame and the train's frame. The coefficient of static friction is provided, which may influence the analysis of forces acting on the passenger.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the passenger in both frames of reference, questioning whether the same forces apply in each case. There is an exploration of the relationship between the forces acting on the passenger and the train's acceleration.

Discussion Status

Some participants have offered insights into the forces that should be represented in the FBDs, including the need to consider pseudo forces in non-inertial frames. There is an ongoing exploration of the correct representation of forces, with no explicit consensus reached on the final approach.

Contextual Notes

Participants are navigating the complexities of analyzing forces in non-inertial frames and the implications of static friction in the context of the problem. The discussion reflects a mix of assumptions and interpretations regarding the forces involved.

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Homework Statement



A passenger is standing without slipping in a forward accelerating train. The coefficient of static friction between the feet and floor is 0.47

a) Draw an FBD for the passenger in Earth's frame of reference
b)Draw an FBD for the passenger in the train's frame of reference.

The Attempt at a Solution



a)

Relative to the earth, we know the passenger in traveling with the same acceleration as the train. Therefore, the train has a force applied and a force of friction in the opposite direction. However, does this also apply to the person who is standing still in the train?

b) For this, I am assuming that there is no force applied or force of static friction because the person is standing still. There is a force normal and force of gravity. Is this right? If it is, do I only draw the FBD with only Fg and Fn?
 
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Won't they be the same? In either frame the man accelerates so there must be a force in the direction of acceleration on the man to cause it to accelerate?
 
Fisics said:

Homework Statement



A passenger is standing without slipping in a forward accelerating train. The coefficient of static friction between the feet and floor is 0.47

a) Draw an FBD for the passenger in Earth's frame of reference
b)Draw an FBD for the passenger in the train's frame of reference.

The Attempt at a Solution



a)

Relative to the earth, we know the passenger in traveling with the same acceleration as the train. Therefore, the train has a force applied and a force of friction in the opposite direction.
The force that is accelerating the train and passengers with respect to the Earth has little to do with the force that is accelerating the passenger with respect to the earth
however, does this also apply to the person who is standing still in the train?
If the person is standing still with respect to the train, then that person must be accelerating with respect to the earth, at the same acceleration of the train with respect to the earth.
b) For this, I am assuming that there is no force applied or force of static friction because the person is standing still. There is a force normal and force of gravity. Is this right? If it is, do I only draw the FBD with only Fg and Fn?
Now the use of non-inertial (accelerating) frames should be avoided like the plague except in certain problems such as this one, since it was asked, so you cannot avoid the question unless you choose not to answer it. FBD's in non-inertial frames require the use of pseudo forces (fictitious forces) and real forces. There is no acceleration of the passenger with respect to the train...since they are both accelerating together...so the real force F is balanced by the make-believe force F' = -ma, where 'a' is the acceleration with respect to the earth. Although they cancel, they both should be shown and properly marked, in the x direction. Fg and Fn still show in either case, since the pseudo force in that direction is 0, so those forces in the vertical direction result in the same FBD of forces in the y direction. Pseudo forces are not fun!
 
Thank you! :)
 

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