Dynamics of block with friction how to set up?

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SUMMARY

The discussion focuses on calculating the force required to push a 2.0 kg wood box down a vertical wall at a constant speed while applying force at a 45-degree angle. The coefficient of kinetic friction (Mk) is established as 0.2 for wood on wood. The correct force (Fa) to maintain constant speed is determined to be 23.1 N, contrasting with the incorrect calculation of 27.71 N. The solution requires a comprehensive analysis of all forces acting in both the x and y directions.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of forces and friction, specifically kinetic friction
  • Ability to resolve forces into components
  • Familiarity with basic algebra for solving equations
NEXT STEPS
  • Study the principles of Newton's second law of motion
  • Learn about friction coefficients and their applications in physics
  • Practice resolving forces into their x and y components
  • Explore static vs. kinetic friction scenarios in physics problems
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of forces and friction in real-world applications.

fizziksc
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Homework Statement


a 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 degree angle. what magnitude of foce should you apply to cause the box to slide down at a constant speed? I know that Mk = .2 (wood on wood)


Homework Equations





The Attempt at a Solution



I tried to solve it with setting the forces in the y direction. I keep getting it wrong. i said Fy=0=Fasin[tex]\vartheta[/tex] - 19.6 N then solved and got Fa = 27.71 N. however the answer is Fa = 23.1 N
 
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fizziksc said:

Homework Statement


a 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 degree angle. what magnitude of foce should you apply to cause the box to slide down at a constant speed? I know that Mk = .2 (wood on wood)


Homework Equations





The Attempt at a Solution



I tried to solve it with setting the forces in the y direction. I keep getting it wrong. i said Fy=0=Fasin[tex]\vartheta[/tex] - 19.6 N then solved and got Fa = 27.71 N. however the answer is Fa = 23.1 N
You have to look at all forces in the y direction. You missed one. You're going to have to look for all forces in the x direction as well.
 

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