What Is the Minimum Angle to Keep the Boxes Stationary on an Inclined Plane?

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SUMMARY

The discussion focuses on calculating the minimum angle α required to keep two boxes stationary on an inclined plane, given their masses of 30kg and 40kg and a coefficient of static friction (μs) of 0.2. The equilibrium equations are applied, leading to the conclusion that the net force must equal zero since the boxes are not accelerating. Participants emphasize the importance of maintaining symbolic representations in calculations until the final step to avoid complications.

PREREQUISITES
  • Understanding of static friction and its coefficient (μs = 0.2).
  • Knowledge of equilibrium equations in physics.
  • Familiarity with trigonometric functions (sine and cosine).
  • Ability to manipulate algebraic equations symbolically.
NEXT STEPS
  • Study the derivation of equilibrium equations for inclined planes.
  • Learn about the effects of varying coefficients of friction on static equilibrium.
  • Explore advanced applications of trigonometry in physics problems.
  • Investigate common pitfalls in algebraic manipulation in physics calculations.
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Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of static friction applications in real-world scenarios.

whitejac
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Homework Statement


9.12)
The mass of the box on the left is 30kg and the mass of the box on the right is 40kg. The coefficient of static friction between each box and the inclined plane is μs = 0.2. Determine the minimum angle α for which the boxes will remain stationary.
20151202_052804.jpg


Homework Equations


μsN = Fs
Equillibrium Equations

The Attempt at a Solution


As you can hopefully see, Box 1 is the yellow box on the right and box 2 is the blue box on the left. My problem is finding N in the second equation. It's just looking too nasty...
Box 1:
Fx = T - Fs - Mgcos(30)
Fy = N - MgSin(30)
Mg = 392.4 => N = 196.2 => Fs39.24 => T = 379.04

Box 2:
Fx = T - Fs - Mgcos(α)
Fy = N - MgSin(α)
Mg = 294.3 => α = Cos-1(N / 294.3)
This answer isn't very useful to me... You can see at the bottom of my notes I tried making substituting alpha into Fx, but it gives a nasty thing to factor. Historically this means I screwed up somewhere rather than the problem being this complicated...?
20151202_052450.jpg
 
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whitejac said:
Box 1:
Fx = T - Fs - Mgcos(30)

I've only have a quick look at this so I might be wrong but I think I can see a few issues with this equation...

1) The box isn't accelerating so what does that say about the net force in any direction?
2) Which way are you assuming friction acts on Box 1?
3) Check you haven't used Cos when it should be Sin?
 
CWatters said:
1) The box isn't accelerating so what does that say about the net force in any direction?
2) Which way are you assuming friction acts on Box 1?
3) Check you haven't used Cos when it should be Sin?
Judging from the working, whitejac has substituted that the net force is zero. But I second your points 2 and 3.
Also, whitejac, I strongly advise against plugging in numerics until the final step. Keep everything symbolic until then. There are many advantages.
 

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