May you me with grade twelve physics?

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The discussion centers on determining the smallest coefficient of friction required to prevent slippage between a small box resting on a larger box, both accelerating together. The acceleration is given as 2/5 m/s², and participants clarify that static friction is the only force preventing the small box from slipping. The relevant equations discussed include Newton's second law and the formula for maximum static friction, leading to the conclusion that the mass cancels out in the calculations. Ultimately, the smallest coefficient of friction is calculated to be approximately 0.26. This approach effectively combines the principles of physics to solve the problem.
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Questions is...
a small is resting on a larger box, which is resting on a table. When an applied force is applied to the larger box, the two boxes move together. The small box does not slip.

If the acceleration of the pair of boxed has a magnitude of 2/5m/s^2, determine the smallest coefficient of friction between the boxes that will percent slippage.

I'm not sure where to even start... If the object is accelerating, wouldn't the small box slip off eventually? Or are we supposed to find static friction? ...
 
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The only force accelerating the small box is static friction. What friction force is required? (Hint: Let "m" stand for the mass of the small box.)
 
Thanks for replying!

Would I need to use a formula where the masses cancel? The only value the question gives us is the acceleration... which is 2.5m/s². Also the 9.8N/kg which is always assumed...

...I'm looking for µ, right?
 
lanvin said:
Would I need to use a formula where the masses cancel?
The mass will cancel, which is why I suggest labeling it "m" and continuing.

...I'm looking for µ, right?
That's right.

Hints: You'll need to combine two formulas: (1) Newton's 2nd law; (2) The formula for maximum static friction. (That second formula contains µ.)
 
alright,
so
F(s) = ma

F(N) - F(g) = 0
F(N) = F(g)
F(N) = mg

F(s) = µ(s) F(N)
ma = µ(s)mg
a/g = µ(s)
[2.5m/s²] / [9.8m/s²]= µ(s)
therefore, the smallest coefficient of friction between the boxes that will prevent slippage is 2.6x10^(-1)

something like that?
 
Exactly like that. :approve:
 
wow thanks
 
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