What Horizontal Force Tips a Crate with Given Friction and Mass?

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SUMMARY

The discussion focuses on calculating the horizontal force required to tip a uniform crate with a mass of 14.1 kg and a coefficient of static friction of 0.551. The crate, a cube with sides measuring 1.21 m, will tip around one of its lower edges when a sufficient horizontal force is applied. The key to solving this problem lies in understanding the balance of torques: the torque from the applied force must equal the torque from the weight of the crate at the tipping point.

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  • Understanding of torque and rotational equilibrium
  • Familiarity with static friction concepts
  • Basic knowledge of forces and mass
  • Ability to apply Newton's laws of motion
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  • Learn about static friction and its role in tipping scenarios
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This discussion is beneficial for physics students, educators, and anyone interested in mechanics, particularly those studying forces and torques in static systems.

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Homework Statement


A uniform crate with a mass of 14.1 kg rests on a floor with a coefficient of static friction equal to 0.551. The crate is a uniform cube with sides 1.21 m in length.
If a horizontal force is applied to the top of the crate, what is the magnitude of the horizontal force needed to cause it to start tipping?
uS=.551
mass=14.1

Homework Equations


t=0

The Attempt at a Solution


I have no idea how to solve this problem,but i know it has something to do with torques. What do i set the axis of rotation to so i can get the torque caused by friction and normal.
 
Last edited:
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Yes, it has something to do with the torque.

Assume the crate is about to tip around one of the four lower edges. Take the torque around that edge. The torque due to the horizontal force has to be balancing the torque due to weight of the crate.
 

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