Extremely frustrated - inclined plane problem w/ friction

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SUMMARY

The discussion focuses on solving an inclined plane problem involving two masses, M1 (6.58 kg) and M2 (7.30 kg), with M1 on a rough incline at an angle of 59 degrees and a coefficient of kinetic friction (μk) of 0.18. The user correctly identifies the need to calculate the frictional force using the equation F_friction = μk * N and applies the principles of Newton's second law to derive the acceleration for both scenarios: M1 moving up and down the incline. The net force equations for both cases are established, leading to a clear method for calculating the acceleration of the system.

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  • Understanding of Newton's laws of motion
  • Knowledge of frictional forces and coefficients
  • Ability to resolve forces into components on an inclined plane
  • Familiarity with basic algebra for solving equations
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  • Learn about the effects of friction on motion
  • Explore advanced applications of Newton's laws in multi-body systems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of inclined plane problems involving friction and acceleration calculations.

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Homework Statement


Mass M1 (6.58 kg) is on a rough inclined plane that makes an angle θ = 59o with the horizontal. The coefficient of kinetic friction between M1 and the incline is μk = 0.18. A rope is attached to M1, passes over a frictionless pulley, and is attached to mass M2 = 7.30 kg which hangs freely.
(a) If M1 is initially moving up the incline: find the magnitude of the acceleration of the masses.
(b) If M1 is initially moving down the incline: find the magnitude of the acceleration of the masses.


Homework Equations


Frictional force = μk * N
F=mg
Components


The Attempt at a Solution


I drew the inclined plane and labeled all of the forces. To solve part (a), M1 is moving up the incline, making the frictional force in the direction of the bottom of the plane, parallel to the plane. The total frictional force should be (μ*N) + (mg*sin(59)), if I am correct. The net force on M1 would be (M2*g) - [(μ*N) + (mg*sin(59))] = (Ʃm)*a

We then divide by Ʃm to get the acceleration.

For part (b), it's the same idea. I would solve it like this:

(mg*sin(59)) - [(M2*g) + (μ*N)] = (Ʃm)*a

We divide by Ʃm again to find the acceleration.

Am I taking the wrong approach? Also, if I'm not being clear on any part, let me know.
 
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If you mean M1 on m your equations are correct. Explain the chosen direction of acceleration in both cases, what direction you take positive.

ehild
 

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