Calculating Forces with Friction and Newton's Laws

AI Thread Summary
The discussion focuses on calculating the forces exerted between two crates using Newton's laws and friction. A friction coefficient of 0.15 is given, with crate 1 weighing 75kg and crate 2 weighing 110kg, under a push of 730N. To find the force each crate exerts on the other, the system's acceleration must first be calculated using F = ma, resulting in an acceleration of 2.48m/s². Free-body diagrams are recommended to analyze the forces acting on each crate, leading to the equations that relate the net forces and the frictional force. The final calculations should incorporate friction to determine the correct forces acting between the crates.
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 F_{12} = -F_{21} (F_{12} - 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 \Sigma F = ma on each box. You know both m and a.

\Sigma F_1 = m_1a

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

\Sigma F_1 = F + F_{21}

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

<br /> \begin{align*}<br /> F + F_{21} &amp;= m_1a \\<br /> F_{21} &amp;= m_1a - F \\<br /> \end{align*}<br />

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 F_12.
 
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is 435 correct then?
 
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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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