Force required to tip block backwards

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SUMMARY

The discussion focuses on calculating the force required to tip a 2-D block backwards when a force F is applied at a height of 0.1H from the ground. The key insight is that the tipping point is influenced by the height of the applied force and the block's dimensions, specifically its height H and width W. The relationship between the applied force, the block's weight, and the resulting torque is critical in determining the tipping condition. The external force creates additional torque, necessitating a greater acceleration for the block to tip backwards.

PREREQUISITES
  • Understanding of torque and its calculation (Torque = Force x Distance)
  • Knowledge of the concepts of static equilibrium and tipping points
  • Familiarity with the effects of friction (coefficient of friction μ)
  • Basic principles of dynamics and acceleration
NEXT STEPS
  • Study the principles of static equilibrium in rigid body mechanics
  • Learn about the effects of different heights of applied forces on tipping points
  • Explore the relationship between friction and tipping stability in blocks
  • Investigate real-world applications of torque in engineering and physics
USEFUL FOR

Students in physics or engineering courses, educators teaching mechanics, and anyone interested in understanding the dynamics of tipping objects and torque calculations.

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Homework Statement


There is a 2-D block with height H, width W, and mass M. (H > W)
A force F is applied from behind the block at a height 0.1H from the ground. If the coefficient of friction is μ, what does the force F need to be to cause the block to tip backwards?


Homework Equations



F x r

The Attempt at a Solution



I know that if the we are interested in tipping the block forwards, then we set Fxr = mgxr and solve for F. Since the question asks for when it tips backward, how do you find the solution?
 
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welcome to pf!

hi simonx! welcome to pf! :smile:

if the block was accelerating with a force applied at ground level, "the weight would gradually shift to the back", ie the reaction force from the ground would get nearer the back

at a certain acceleration, the reaction force reaches the back, so there is no longer any torque to keep the box from rotating, and the block is free to tip

here, the external force is at height H/10, so that's an extra torque, and the tipping acceleration must therefore be greater :wink:
 

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