Car moving with two blocks attached to it.

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Homework Help Overview

The discussion revolves around a physics problem involving a car with two blocks attached, focusing on the forces required to keep the blocks stationary relative to the car. The scenario assumes no friction in any surfaces, pulleys, or wheels.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of frictionless surfaces on the forces acting on the blocks and the car. Questions arise regarding the relationship between the applied force and the tension in the rope connecting the blocks. There is discussion about analyzing the system as a whole versus using free body diagrams to understand internal and external forces.

Discussion Status

Participants are actively engaging with the problem, sharing insights and clarifying concepts. Some have provided guidance on understanding the tension required to keep the blocks stationary and the implications of different approaches to solving the problem. There is recognition of the complexities involved in analyzing the system.

Contextual Notes

Participants note the challenge of reconciling different methods of analysis, particularly when considering the effects of the blocks on the car's motion. The discussion reflects on the assumptions made regarding the forces at play and the resulting equations derived from those assumptions.

ShizukaSm
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Homework Statement


Car.PNG

What force must be applied in the car of the image in order to all blocks remain stationary in relation to the car? Supposed that there's no friction in all surfaces, pulleys and wheels.

Homework Equations



Fr = ma

The Attempt at a Solution


First thing: That's a doubt I have, since there is no friction, the force F applied on the cart won't actually affect the block m1 in anyway, right? I mean, if there was no m2, the block m1 would stand in the air while M rushed in a straight line, right?

Ok, so I made all the equations and free body diagrams:

For block m1:
[itex]T = m_1a[/itex]

For block m2(F_32 is the force that the car M exerts on block with mass m2, by the way):
[itex]\\T - P_2 = 0\\<br /> F_{32} = m_2a[/itex]

And finally, for the car(Again, F_23 is the reaction to F_32):
[itex]F - F_{23}= Ma[/itex]

Substituting everything, I find:
[itex]F = \frac{m_2g(m2+M)}{m1}[/itex]

The answer, however, is:
[itex]F = \frac{m_2g(m1+m2+M)}{m1}[/itex]

Right, so, the thing is, I've learned that there are mainly two ways of working this kind of problem:
The first one is to look at the whole system, and only analyze the external forces. The downside is that I won't know the internal forces.
The second one is to look at each free body diagram and calculate the relevant forces. The downside being that it takes more time to do in this way.

That being said, both ways always give the same answers, they're just different paths.

However, in this case, I've noticed that if I consider the system as a whole, I will get the right answer (Since mtotal = m1 + m2 + M), when considering the free body diagrams though, the block 1 won't act on car M, so he doesn't get in.

Where am I wrong?


Thanks in advance!
 
Last edited:
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ShizukaSm said:

Homework Statement


View attachment 54666
What force must be applied in the car of the image in order to all blocks remain stationary in relation to the car? Supposed that there's no friction in all surfaces, pulleys and wheels.

Homework Equations



Fr = ma

The Attempt at a Solution


First thing: That's a doubt I have, since there is no friction, the force F applied on the cart won't actually affect the block m1 in anyway, right? I mean, if there was no m2, the block m1 would stand in the air while M rushed in a straight line, right?
That's true, so the idea is to make sure that the tension in the rope just keeps m2 from falling. Then that tension will also keep m1 stationary with respect to the cart.

What tension in the rope will keep block m2 from falling? (What tension balances the vertical components of its FBD?)

What acceleration of m1 will produce this tension? (think inertial force)
 
Right! But in fact, I did that. P2 = T means it will remain stationary (I actually wrote that as P2 - T = 0, but it's the same thing of course). I also found the acceleration correctly.

My biggest problem was, as I said, the last part when analyzing the big cart alone.
 
The string presses the pulley: it is attached to the cart, so the cart experiences force from the tension in he string.

ehild
 
Wow, that is... amazing, and does make sense. That is truly amazing, thanks a lot!
 
You are welcome :)

ehild
 

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