Determine if the block will slide or flip over after contact

  • Context: Undergrad 
  • Thread starter Thread starter vcl0124
  • Start date Start date
  • Tags Tags
    Friction Sliding
Click For Summary

Discussion Overview

The discussion revolves around determining whether a block (C) will slide or flip over when a force is applied to another block (B) that is fixed in the Y direction and in contact with a fixed block (A). The scenario involves analyzing the forces and moments acting on the blocks, with known parameters such as angle (theta), height (h), width (w), and mass (m). Participants explore the conditions under which sliding or tipping occurs, as well as the possibility of the blocks locking up.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines a strategy involving force balance equations and the relationship between static and kinetic friction to determine the conditions for sliding or tipping.
  • Another participant seeks clarification on how the applied force reaches block C and whether block B is sliding under block A.
  • It is noted that the increasing force applied to block B will cause it to slide, which in turn affects block C due to friction.
  • A participant suggests considering the possibility of the blocks locking up, making motion impossible until a failure occurs.
  • Several participants discuss the implications of different angles (theta) and friction coefficients on the sliding or tipping behavior of block C.
  • One participant provides a detailed approach using a free body diagram (FBD) to analyze the forces and moments, suggesting calculations for various cases based on different angles and friction coefficients.
  • Another participant shares their calculations based on the suggested method and discusses the implications of their results regarding sliding and tipping conditions.

Areas of Agreement / Disagreement

Participants express varying views on the conditions for sliding versus tipping, with some proposing calculations and scenarios while others suggest alternative possibilities like locking up. There is no consensus on a definitive outcome, and multiple competing views remain.

Contextual Notes

Participants mention several assumptions and parameters that influence the analysis, including the coefficients of friction and the angle of inclination. The discussion reflects uncertainty regarding the exact conditions under which sliding or tipping will occur, as well as the potential for locking up.

vcl0124
Messages
8
Reaction score
0
Untitled.png

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.
 
Physics news on Phys.org
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?
 
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
 
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?
 
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.
 
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)?
 
  • Like
Likes   Reactions: Lnewqban
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.
 
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.
 
  • Like
Likes   Reactions: Lnewqban
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
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 112 ·
4
Replies
112
Views
8K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K