Person in a bus (accelerating system)

[pinsky]

Hy

I'm having some issues about the Newtons third law.

The image should represent a person in a bus with and the forces acting on them.
The bus and the person are observed from outside the bus. The bus starts accelerating because of the force $F_{BUS MOVING}$. This force represent the resultant force of the friction between the bus and the ground, and the buses starting force.


As a result, friction force $F_{P-BUS}$ occurs and it acts on the bus because of the mass of the person.
As a reaction, a force of the same size but opposite direction acts on the person $F_{BUS-P}$

What is the sum of the forces on the bus?
$F_{BUS MOVING}-F_{P-BUS}$ ?

How would one calculate $F_{P-BUS}$ by knowing the mass of the person and $\mu$?

Is this the solution?
$F_{BUS MOVING}-F_{P-BUS}=F_{BUS-P}$

Since in this case, the bus and the person have the same value and orientation. That is what a observer from the street would see, right?

 

[tiny-tim]

Welcome to PF!

Hi pinsky! Welcome to PF!

(try using the X₂ tag just above the Reply box )

F_P-BUS always equals (minus) F_BUS-P.

If you regard the friction as an internal force, then you ignore it, and F = (M_BUS + m)a.

If you regard the friction as an external force, then you include it, and F - F_P-BUS = M_BUSa.

 

[pinsky]

Thanks for the welcome.

I didn't understand your explanation though.

Could you perhaps draw all the forces in a situation where the bus is starting, and the friction between the bus and the person in the bus isn't enough to prevent the person from gently sliding. So in that case we can't observe the bus and the person as a single object.
Draw the forces as they appear from the point of view of a person standing still outside the bus.

I made you a template to spare the time :)

 

[tiny-tim]

I'm not drawing it, but if the force on the bus is F, and the friction force is G, then

F - G = M_BUSa_BUS,

G = m_Pa_P

 

[PJC01]

I can provide a response to this content by explaining the concept of Newton's third law and how it applies in this scenario.

Newton's third law states that for every action, there is an equal and opposite reaction. In this case, the action is the force of the bus accelerating, and the reaction is the force on the person.

The sum of the forces on the bus can be calculated by adding all the external forces acting on the bus, including the force of the bus accelerating and the friction force between the bus and the ground. This sum of forces will determine the acceleration of the bus.

To calculate the friction force F_{P-BUS}, we can use the formula F_{P-BUS} = \mu * m_{person} * g, where \mu is the coefficient of friction between the bus and the ground, m_{person} is the mass of the person, and g is the acceleration due to gravity. This force is equal in size but opposite in direction to the force on the person, F_{BUS-P}.

Therefore, the solution provided in the content is correct. The sum of the forces on the bus is F_{BUS MOVING}-F_{P-BUS}=F_{BUS-P}, and this is what an observer from the street would see. Both the bus and the person experience equal and opposite forces, resulting in the bus accelerating forward while the person appears to stay in place.

It is important to understand Newton's third law to accurately analyze and calculate forces in a system, especially in cases where there is acceleration involved. I hope this helps clarify any confusion you may have about the concept.