Calculating Forces with Friction and Newton's Laws

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Homework Help Overview

The discussion revolves around calculating the forces between two crates in the context of Newton's Laws of motion, specifically considering the effects of friction. The original poster is seeking assistance with a problem involving a push force and the masses of the crates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss treating the system as a whole to find acceleration and suggest using free-body diagrams. The original poster expresses confusion about the relationship between acceleration and the forces exerted between the crates.

Discussion Status

The conversation includes attempts to clarify the problem and provide guidance on how to approach the calculations. Some participants have offered methods to find the forces acting on the crates, while the original poster continues to seek clarity on the next steps.

Contextual Notes

There is mention of a friction coefficient and the need to incorporate it into the calculations, indicating that the problem may have additional complexity. The original poster has not provided complete details of the scenario, which may affect the discussion.

aceXstudent
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Hi there,

I'm having some trouble with this problem right now. We are learning about Newton's First/Second/Third Laws.

With a friction coefficient of 0.15, what is the force that the crates exert on each other if the crate 1 is 75kg and crate 2 is 110kg with a push of 730N?

Can anyone give me pointers or starters?
 
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Treat the system as a whole to get acceleration.
 
Errrr... I was looking for the force that each box exerts on each other, not the acceleration...

But thanks anyways.
 
aceXstudent said:
Errrr... I was looking for the force that each box exerts on each other, not the acceleration...

But thanks anyways.

I didn't give you the complete story. You asked for pointers/starters.

Let me elaborate: Treat the system as a whole to find the acceleration of the entire system (use F = ma.) The system moves as a whole, thus each box will have the acceleration that you found. Continue by drawing free-body diagrams for each box and keep in mind that [itex]F_{12} = -F_{21}[/itex] ([itex]F_{12}[/itex] - read "force acted on box two by box one.")
 
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-_-, still confused, sorry... acceleration is 2.48m/s^2-- what do i do with that number?
 
aceXstudent said:
-_-, still confused, sorry... acceleration is 2.48m/s^2-- what do i do with that number?

I don't know if you read my edited post but use [itex]\Sigma F = ma[/itex] on each box. You know both m and a.

[tex]\Sigma F_1 = m_1a[/tex]

You now have the net force on box 1. If you drew your free-body diagrams you would also know that

[tex]\Sigma F_1 = F + F_{21}[/tex]

Our last two expressions are both give us the net for on box 1 so we can equate them

[tex] \begin{align*}<br /> F + F_{21} &= m_1a \\<br /> F_{21} &= m_1a - F \\<br /> \end{align*}[/tex]

Tada! You have the force acting on box 1 by box 2. If you read what I said in my last post you should be able to get [itex]F_12[/itex].
 
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is 435 correct then?
 
Last edited:
Looking back, I see your problem includes friction. It shouldn't be too hard for you to include that in with what I've given you. Recheck your calculation of acceleration too.
 

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