[Friction] Angled bar pressing down against a plate

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AI Thread Summary
The discussion revolves around solving a physics problem involving friction and forces on a bar pressing against a plate. The initial approach correctly identifies the angle alpha using the equation tan(alpha) = f/N, leading to alpha = arctan(u). However, confusion arises in the second part regarding the sum of moments, where an incorrect setup initially leads to an infinite friction force. After clarification and adjustments to the equations, the correct friction force is determined to be F = uW/4, resolving the misunderstanding. The conversation highlights the importance of accurately setting up equations for different parts of the problem.
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Homework Statement


VIfUiu2.png

This is my problem in all its glory.

Homework Equations


f = uN

The Attempt at a Solution


For part a I simply did that tan(alpha) = f/N and therefore alpha = arctan(f/N).
This simplifies to alpha = arctan(u). I believe this part is correct.

The second part of this question is what gets me. I set up the sum of the moments for the bar.
My equation is: W*sin(alpha)*L/2 - N*sin(alpha)*L + f*cos(alpha)*L = 0

Algebra gave me that N = - W*sin(alpha) / [2*(u*cos(alpha) - sin(alpha)]

I made a FBD for the plate and the equations I got were that F - f = 0 and Ngroundonplate - N = 0. (The plain N being the normal force of the bar on the plate)

Since f = uN, substitution of N gave me f = - u*W*sin(alpha) / [2*(u*cos(alpha) - sin(alpha)] and then dividing top and bottom by cos(alpha) I got that f = INFINITY.

If this makes sense, can anyone explain how? If it's wrong can anyone point me in the right direction?
 
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bartersnarter said:
W*sin(alpha)*L/2 - N*sin(alpha)*L + f*cos(alpha)*L = 0
Which way is f acting on the bar?
 
If the workpiece is moving to the left, the friction force should be towards the right, shouldn't it?
 
bartersnarter said:
If the workpiece is moving to the left, the friction force should be towards the right, shouldn't it?
Is that the friction force on the workpiece or the friction force on the bar?
 
The friction on the workpiece.
 
bartersnarter said:
The friction on the workpiece.
Ok, but the equation I quoted is in regard to forces on the bar, no?
 
Ahh, I see now! I had set up that sum of moments for part a and I neglected to alter it for part b. Thanks a bunch!
After altering it I get F = uW/4. This makes much more sense!
 
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