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

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To calculate acceleration on an inclined plane with friction, the relevant equation is gsinθ - ugcosθ = a, where g is gravitational acceleration, θ is the angle of incline, and u is the coefficient of friction. The box starts sliding at an angle of 36.8699 degrees, with a coefficient of kinetic friction of 0.4, leading to a calculated static friction coefficient of 0.75. The user attempted to solve for acceleration but received a different answer than the expected 2.744, indicating a misunderstanding of the friction type applicable once the box is in motion. It is crucial to differentiate between static and kinetic friction when analyzing the problem. Understanding these concepts will help clarify the calculations involved.
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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