Minimum Coefficient of static friction to keep a block from sliding?

AI Thread Summary
To determine the minimum coefficient of static friction required to prevent a 5kg block from sliding on a 15kg block under a 40N horizontal force, the system must be analyzed for acceleration. Initially, the calculation incorrectly assumed the blocks would accelerate together, leading to a coefficient of 0.20. However, the correct approach involves recognizing that the frictional force must counteract the applied force, resulting in the equation (5kg)(2m/s^2) = μFN + 40N. After correcting the calculations, the minimum coefficient of static friction is found to be approximately 0.61. This highlights the importance of accurately accounting for all forces in friction-related problems.
Brodo17
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A 5kg block is placed on top of a 15kg block that rests on a frictionless table. The surface between the top and bottom blocks is roughened so that there is no slipping between the blocks. A 40N horizontal force is applied to the top block. What is the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block?


F=ma
f=mew FN


Ok, in my first attempt I figured that if the blocks don't slide apart, then the entire sysstem would be accelerating. I said 40N = (20)a, and found that a=2m/s^2
Using this I said (5kg)(2m/s^2) = mew FN
10 / (9.8 X 5) = .20

However the answer says that it is actually .6

Then I though that maybe the system could not accelerate at all because the opposing frictional force will stop it from accelerating.
Im not even sure if I am on the right track though, and I cannot figure this out!

Thank you!
 
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"Using this I said (5kg)(2m/s^2) = mew FN"

You forgot the 40 N pulling force.

ehild
 
What do you mean I forgot about the 40N force? Where would I put it in my equation?

Actually, If I go
(5kg)(2) = mewFN + 40N
-30N = mewFN
-30 / (-9.8 x 5) = .61

Thanks
 
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