Coefficient of Friction and Forces

AI Thread Summary
To determine the coefficient of friction (μ) between the box and the flatbed trailer, the discussion highlights the need to calculate the acceleration (a) using the final velocity and distance traveled. The formulas FF = μFN and FN = mg are mentioned, but the lack of mass complicates the calculations. Participants suggest that writing out the equations may allow for mass to cancel out, simplifying the process. Kinematic equations are referenced as a potential resource for finding acceleration. Ultimately, the conversation revolves around finding the necessary values to solve for μ effectively.
pippintook
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A flatbed tractor trailer takes off from a stop sign on a level road. If a box on the back slides back 10 meters while the truck goes forward 50 meters and attains a speed of 15 m/s, what is the coefficient of friction (μ) between the bed and the box? Assume the bed to be at least 10 meters long.


FF= μFN
FN= mg


I don't know if those are the right formulas, specifically because no mass is given. I just don't know how to find the forces without acceleration; is there a formula for that? If these are the right formulas, how do I find the mass or FN? I chose these formulas because FF is the forward force, and it had μ in the equation. Help!
 
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That's some of them.

There is also F = m*a.

And that a would be really useful to know wouldn't it?

You know the final velocity and you know the distance, so you should be able to figure the "a" shouldn't you?
 
Wow. Ok, so I've found "a", but still don't know how to get FN, since I don't have the mass or force. I know there's some really obvious formula, but I CANNOT remember it.
 
pippintook said:
Wow. Ok, so I've found "a", but still don't know how to get FN, since I don't have the mass or force. I know there's some really obvious formula, but I CANNOT remember it.

Maybe write out your equations and your mass will cancel out?

If you are looking for kinematic equations:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2
 
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