Concept Problem: Oscillation and friction

[cjavier]

So I feel as though I have the correct solution, but am not positive. My problem is as follows: A block of mass M is at rest with respect to a surface which oscillates horizontally with sinusoidal motion described by the equation x(t)=Asin(ωt). Find an expression for the minimum value of the coefficient of static friction between the block and surface so that the block does not slip as the surface oscillates. ALSO: If the block were to slip, where in the oscillation would this happen?

I stated that if the block were to slip, it would be when x''(t) is at a maximum. The expression for x''(t) is -ω²Asin(ωt) and so when ωt=^∏/₂, the acceleration is at it's max, because the sine is at 1.

And I set the kinetic force equal to this expression for acceleration times the mass of the block. Like so: Mgμ_k = -ω²A.

I finished with μ_k = ^-ω2A / _g
Let me know if you feel as though I am correct. Thanks!

 

[TSny]

Overall, that looks very good. But are you dealing with kinetic friction or static friction?

(In your force equation, I think you left out the mass on the right side. But, I believe that's just a typing error.)

 

[cjavier]

Overall, that looks very good. But are you dealing with kinetic friction or static friction?

(In your force equation, I think you left out the mass on the right side. But, I believe that's just a typing error.)

Static friction, and yes I did, thank you

 

[TSny]

The coefficient of friction should be a positive number. So, you should consider how you ended up with a negative value.

 

[Nickie]

Your solution seems to be correct. To find the minimum coefficient of static friction, we need to consider the maximum acceleration of the surface, which occurs when the sine function is at its maximum value of 1. This corresponds to when ωt=∏/2, as you stated.

To prevent the block from slipping, the maximum static friction force must be equal to or greater than the maximum kinetic force. This can be expressed as μsN≥μkN, where μs is the coefficient of static friction, N is the normal force (equal to the weight of the block, Mg), and μk is the coefficient of kinetic friction.

Substituting in the values for N and the maximum kinetic force (-ω2A), we get μsMg≥(-ω2A)M. Simplifying, we get μs≥-ω2A/g.

So your expression for the minimum coefficient of static friction is correct: μs≥-ω2A/g. This means that the coefficient of static friction must be greater than or equal to this value in order for the block to not slip as the surface oscillates.

As for where in the oscillation the block would slip, it would occur at the point where the coefficient of static friction is equal to the minimum value we calculated. This would be when ωt=∏/2, or at the maximum displacement of the surface. At this point, the static friction force would be exactly equal to the maximum kinetic force, and any further increase in acceleration would cause the block to slip.

Overall, your solution and reasoning are correct. Good job!