How Do You Calculate Acceleration on an Inclined Plane with Friction?

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SUMMARY

The discussion centers on calculating acceleration for a box sliding down an inclined plane with friction. The coefficient of kinetic friction is 0.4, and the incline angle is 36.8699 degrees. The user successfully determined the coefficient of static friction to be 0.75 using the equation Mk = tan(θ). However, confusion arises regarding the application of the equation gsin(θ) - ugcos(θ) = a, leading to a discrepancy between the calculated acceleration and the expected value of 2.744 m/s².

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with friction coefficients (static and kinetic)
  • Knowledge of trigonometric functions in physics
  • Ability to manipulate equations involving acceleration and forces
NEXT STEPS
  • Review the concept of static vs. kinetic friction in physics
  • Learn how to apply Newton's second law to inclined planes
  • Study the derivation and application of the equation gsin(θ) - ugcos(θ) = a
  • Explore examples of acceleration calculations on inclined planes with varying friction coefficients
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This discussion is beneficial for physics students, educators, and anyone studying mechanics, particularly those focusing on inclined planes and frictional forces.

alex_todd
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Homework Statement


he question is:
A box is sitting on a board. The coefficient of kinetic friction between the box and the board is 0.4. One side of the board is raised so that the board is 36.8699 degrees from horizontal. This is the angle that the box starts sliding.There are two parts to the question, I figured out part A (What is the coefficient of static friction between the box and the board?) by setting Mk equal to Tan and got .75

The part that throws me off about this equation is the last sentence " This is the angle that the box starts sliding"
I'm pretty sure that if it weren't for that, I would use the equation gsinθ - ugcosθ = a however I tried that and didn't get the right answer

Homework Equations


mgsinθ - umgcosθ = ma
gsinθ - ugcosθ = a
mk=tan theta

The Attempt at a Solution


I tried plugging the values into the gsinθ - ugcosθ = a equation, I assumed I could just get rid of the mass variables because they'll all cancel out anyways
gsinθ - ugcosθ = a = 9.8sin(36.8999 degrees) - .75(9.8)cos(36.8999 degrees)
the answer I got doesn't my teachers solution of 2.744
I'm think the part that's throwing me off is "this is the angle that the box starts sliding)
 
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Hi alex_todd, Welcome to Physics Forums!

alex_todd said:
I tried plugging the values into the gsinθ - ugcosθ = a equation, I assumed I could just get rid of the mass variables because they'll all cancel out anyways
gsinθ - ugcosθ = a = 9.8sin(36.8999 degrees) - .75(9.8)cos(36.8999 degrees)
the answer I got doesn't my teachers solution of 2.744
I'm think the part that's throwing me off is "this is the angle that the box starts sliding)
Once the box is moving, what type of friction applies?
 

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