Determine if the block will slide or flip over after contact

In summary, determining whether a block will slide or flip after contact involves analyzing the forces acting on it, including friction, gravity, and the angle of contact. Factors such as the block's weight, surface texture, and the point of application of force are crucial in predicting its motion. By assessing these variables, one can conclude whether the block will slide across the surface or rotate about its edge.
  • #1
vcl0124
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I have a problem like above. A is fixed, B is fixed in Y direction and C is placed on B. Theta, h, w and m are known input parameters. If an increasing force is applied on B towards -ve x direction, I want to determine whether C is slide or flip over. Before slide or flip happened, I believe I have below force balance equations.
Untitled.png

uA and uB are used to represent the static friction force before sliding happened and they should be smaller than uA,k and uB,k (kinetic friction coefficient). When the applied force increases, uB increases and the block start to slide when uB=uB,k. x represents the location of NB. My understanding is when x=0, the block start to flip. My strategic is calculate x as a function of uB and see which one reached the criteria first. However, there are 4 unknown (uA, NA, NBand x) but I only got 3 equations. So my first question is whether my equations and strategic are correct? Moreover, what do I missed to solve this problem?

PS. it is not homework, please don't give me another warning.
 
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  • #2
Welcome! 😎

Could you please explain this part a little more?
"If an increasing force is applied on B towards -ve x direction"

How is that increasing force reaching C?
Is B sliding under fixed A from left to right?
 
  • #3
How is that increasing force reaching C? --> Friction between B and C
Is B sliding under fixed A from left to right? --> Yes, and there is no friction between A and B
 
  • #4
Thank you.
The location of application of Fb should move towards the right bottom edge after the movement is initiated, making the block like an extension ladder supported by an inclined wall.
Will the sliding velocity of B steadily increase due to the increasing (magnitude of that) force?
 
  • #5
Will the sliding velocity of B steadily increase due to the increasing (magnitude of that) force?
This is not what I interest, I only interest on whether C will slide (against B) or flip over first. Or I should say sliding or flip over requested a smaller force on B.
 
  • #6
vcl0124 said:
I only interest on whether C will slide (against B) or flip over first.
Have you considered the third possibility, that it will lock up, making motion impossible (until something breaks)?
 
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  • #7
A.T. said:
Have you considered the third possibility, that it will lock up, making motion impossible (until something breaks)?
Yes, especially when theta is small.
 
  • #8
Here's how I would solve this problem. Start with a free body diagram (FBD) for the case of incipient tipping (just starting to tip). The block is supported at the bottom right edge with a vertical force. There is also a horizontal force due to the friction from block B, which is sliding from right to left. There is a normal force due to contact with part A. That force is normal to the surface of part A. The block is starting to tip, so the contact edge is sliding against part A, and there is a friction force parallel to the surface of part A. This is shown in the FBD below.
FBD block.jpg

Start the solution by taking the sum of moments about the bottom right edge, and solve for force ##F_A##. Then calculate ##F_B## and ##\mu F_B## from the sum of forces in both X and Y directions. Then calculate ##\mu## from ##F_B## and ##\mu F_B##. That should be the limiting value of ##\mu## at which it transitions from sliding to tipping. Smaller ##\mu## will slide, larger will tip.

Run some test cases to confirm your results.
Case 1: Angle ##\theta## = 90 degrees, and ##\mu## A = 0. This is a simple tipping problem, slightly complicated because the tipping force is from friction and the weight of the block C.
Case 2: Angle ##\theta## = 90 degrees, and ##\mu## A = 1. It should require more friction from part B to tip.
Case 3: Angle ##\theta## = 45 degrees, and ##\mu## A = 0. I think this angle will not tip, but check.
 
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  • #9
jrmichler said:
Here's how I would solve this problem. Start with a free body diagram (FBD) for the case of incipient tipping (just starting to tip). The block is supported at the bottom right edge with a vertical force. There is also a horizontal force due to the friction from block B, which is sliding from right to left. There is a normal force due to contact with part A. That force is normal to the surface of part A. The block is starting to tip, so the contact edge is sliding against part A, and there is a friction force parallel to the surface of part A. This is shown in the FBD below.
View attachment 349246
Start the solution by taking the sum of moments about the bottom right edge, and solve for force ##F_A##. Then calculate ##F_B## and ##\mu F_B## from the sum of forces in both X and Y directions. Then calculate ##\mu## from ##F_B## and ##\mu F_B##. That should be the limiting value of ##\mu## at which it transitions from sliding to tipping. Smaller ##\mu## will slide, larger will tip.

Run some test cases to confirm your results.
Case 1: Angle ##\theta## = 90 degrees, and ##\mu## A = 0. This is a simple tipping problem, slightly complicated because the tipping force is from friction and the weight of the block C.
Case 2: Angle ##\theta## = 90 degrees, and ##\mu## A = 1. It should require more friction from part B to tip.
Case 3: Angle ##\theta## = 45 degrees, and ##\mu## A = 0. I think this angle will not tip, but check.
Thanks a lot for your reply! I am sorry that I was busy last week so no reply to you. I followed you suggestion with following assumptions, θ=85°, w=1, h=1, mg=1, μAA,k=0.3 and μB,k=0.5. Following your calculation, I got...

FA=mgw/2[(sinθ-μAcosθ)h-(cosθ+μAsinθ)w]=0.856
FB=FA(cosθ+μAsinθ)+mg=1.330
μB=(FAsinθ-μAcosθ)/FB=0.624

So μBB,k. I think it means tipping won't happen, am I correct? As θ is close to 90°, I believe it is likely slide rather than lock up (the third case). For the sliding case, following your idea, I think we can set μBB,k=0.5 and I got...

μA=[FAB,kcosθ-sinθ)-μB,kmg]/[FAB,ksinθ+cosθ)]

Below is the μA vs FA plot. As |μA|≤μA,k=0.3, I calculated 0.443≤FA≤0.643 is needed. I guess it means sliding happens when FA reaches 0.443, am I correct?

Untitled.png
 
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