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A block hits a patch of friction

  • Thread starter stormgren
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  • #1
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[SOLVED] A block hits a patch of friction

A block of uniform density, length L and height h << L, starts from rest at the top of a hill of height H, slides down. At the bottom there is a flat surface of length greater than L, and then a rough patch with sliding coefficient [tex]\mu[/tex]


I figured out the first part, which is to find the height such that the block comes to a stop the instant the entire block is on the rough surface. The answer is [tex]Hcrit=\frac{L\mu}{2}[/tex] Is this right?

Now I need to find what happens if the hill has a height H < Hcrit.
I think the block still has follows simple harmonic motion, and obviously from conservation of energy it has a speed [tex]\sqrt{2gH}[/tex] at the bottom of the hill, but for some reason I'm having huge trouble finding the distance that the leading edge of the block travels across the surface with friction before it comes to a stop.
I don't know where to go from here. Some hints, please? :)
 

Answers and Replies

  • #2
Shooting Star
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How did you get the answer to the first Q. Did you take the average frictional force? If, so, you have to do the 2nd in the same way.

Suppose it slides upto a dist x in the patch. Take d as the linear mass density. Repeat what you've done for Q1.

[The answer will be x=sqrt(2Hl/mu).]
 
  • #3
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I did the first part like this:

Ffric = (mu)*Fn where Fn is the normal force of the portion of the block, x, on the rough surface.

So Ffric=(mu)*mg/L*x

-mg(mu)*x/L = m*(x'') <---- second deriv of x

So the x(t) equation satisfies simple harmonic motion a= -omega^2*x

x(t) = Lsin(sqrt((mu)*g/L)t)

Then use cons. of energy.


I thought to solve the second part the same way, but perhaps using an average frictional force is better. How do I do that?
 
  • #4
Shooting Star
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There is no simple H motion involved here. It is all about sliding friction.

The block has a certain KE when it hits the rough patch. At the first point, the friction is zero. In between, only a part of it exerts normal force on the patch, the part which has slid over the patch. As it moves more over the patch, the N on the patch increases and so does friction F. When it stops just after covering its own length, the block exerts the whole weight on the patch, and the friction F is max. Instead of integrating over this path, I thought you can take the average frictional force, which is half of F_max+F_min.

Now go back to Q1.
 
  • #5
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I know there's no simple harmonic motion per se, but you still get

[tex]F_{net}=ma[/tex]
[tex]-F_{fric}=m\frac{d^{2}x}{dt^{2}}[/tex]
[tex]-\frac{mg\mu}{L}x=m\frac{d^{2}x}{dt^{2}}[/tex]
[tex]-\frac{g\mu}{L}x=\frac{d^{2}x}{dt^{2}}[/tex]
Which still satisfies the SHM diff eq. That was how we were taught to solve a similar problem, but it as you point out it doesn't seem to be the best way. I'll try it using an average force and see what happens.
 
Last edited:
  • #6
Shooting Star
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OK, since the frictional force increases with x as the block slides more into the patch, you get the dame diff eqn as SHM. Then you can apply it for the 2nd case too. Only the max displacement is less than L this time.
 
  • #7
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Okay, cool... I'm in agreement with your answer of sqrt(2HL/mu).

Thanks for your help! I don't know why I didn't see it before.
 

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