Solving the Mystery of Standing Still in a Moving Train

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The discussion focuses on analyzing the forces acting on a passenger standing in a forward-accelerating train. In Earth's frame of reference, the passenger experiences both the force of friction opposing the train's acceleration and the force of gravity. In the train's frame, the passenger is considered to be at rest, leading to the conclusion that only normal and gravitational forces need to be depicted, along with the inclusion of pseudo forces to account for the acceleration. The importance of correctly representing forces in both inertial and non-inertial frames is emphasized, particularly in drawing free body diagrams (FBDs). Understanding these forces is crucial for solving problems related to motion in accelerating frames.
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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! :)
 
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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