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Suppose I have a mass sitting on a surface with friction. Now I start pushing on the mass in one direction (call this direction the x direction). To get this mass accelerating, I have to push the mass with a force greater than $\mu_{s}mg$. Assuming the force is large enough such that the mass moves until there is a significant x component of velocity.

Now here's where my question arises. While this mass is moving in the x direction suppose I decide to push it along a direction orthogonal to it's motion (call this direction y). Is the minimum force required to get the mass to move in the y direction going to be $\mu_{s}mg$ or is it $\mu_{k}mg$ since the object is already moving?

How would I think of this problem in terms of a microscopic model of friction?

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There doesn't seem to be a clear and obvious answer from first principles, but if the object is in motion, kinetic friction should be applied. It might be useful to consider the microscopic model - that the static form of friction is applied when there are molecular attractions between the object and the floor.

However, kinetic friction deals with dynamic forces that are random and time dependent. This can be due to how the floor is not perfectly uniform, or that there are imperfections in the object. Then, kinetic friction is somewhat an average of many different electromagnetic interactions between the object and the floor.

What you are describing seems to be much more kinetic than static. The object is not at some equilibrium with the floor - it is passing through many different patches. Therefore, it is interacting with the floor in a different way than static friction implies.

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If you are applying two orthogonal forces their resultant (Which is always greater than both the constituent forces) should be greater than static friction or:

Lets say your forces are $ F_x $ and $ F_y$

The magnitude of these forces' resultant is $ \sqrt{(F_x)^2 + (F_y)^2} $

For the block to move the resultant must be greater than friction

$ \sqrt{(F_x)^2 + (F_y)^2} > \mu_smg$

However in your example since the block is already moving any amount of force you produce on the y axis will always increase the magnitude of the resultant force. and hence all the while you will use $\mu_k$ mathematically:

$ F_x > \mu_smg $

$ (F_x)^2 > (\mu_smg)^2 $

$ F_x^2 + F_y^2 > (\mu_smg)^2$

and therefore

$ \sqrt{(F_x)^2 + (F_y)^2} > \mu_smg$

And since $\mu_s > \mu_k $

$ \sqrt{(F_x)^2 + (F_y)^2} > \mu_kmg$

And notice $F_x$ itself is sufficient to get the block move and hence you use $\mu_k$ So while we work with the resultant which is obviously greater than $F_x$ this time too we use $ \mu_k$ because in this case too friction proves insufficient tto stop motion of block

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