Angle required to keep block from sliding down plane

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SUMMARY

The discussion focuses on determining the angle required to prevent blocks from sliding down an inclined plane. The user initially applied the kinetic friction coefficient (μ_k) instead of the static friction coefficient (μ_s), leading to confusion in their calculations. Key equations include F = ma and F_k = μ_kF_N, with the user attempting to balance forces but misapplying the normal force (F_N) and friction concepts. The correct approach involves using static friction to analyze the forces acting on the blocks at rest.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to resolve forces into components on an inclined plane
  • Familiarity with basic algebra and trigonometry
NEXT STEPS
  • Learn about static friction and its application in equilibrium problems
  • Study the concept of force components on inclined planes
  • Explore the differences between static and kinetic friction coefficients
  • Practice solving equilibrium problems involving multiple blocks on inclined surfaces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and friction, as well as educators looking for examples of static versus kinetic friction in problem-solving scenarios.

leroyjenkens
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Homework Statement



Here's a picture of the set up.

I need to find out what angle I need to have the plane raised to so that the blocks aren't moving.


Homework Equations



F = ma

The Attempt at a Solution


For the first block
[tex]F_k=μ_kF_N[/tex]
[tex]F_T+F_N-μ_kF_N+2mgcosθ+2mgsinθ=0[/tex]

Since the [tex]F_N[/tex] and [tex]2mgcosθ[/tex] cancel each other, I eliminated them from the problem. I set the right side equal to 0 since there is no acceleration

Then I get
[tex]F_T = μ_kF_N + 2mgsinθ[/tex]

For the right block [tex]mg = F_T[/tex]
So I input that into my equation for the other block and get
[tex]mg=μ_kF_N+2mgsinθ[/tex]

For [tex]F_N[/tex] I plug in [tex]2mgcosθ[/tex]

And get [tex]mg=μ_k2mgcosθ+2mgsinθ[/tex]

And this is when I know something is wrong. That [tex]μ_k[/tex] is the coefficient for kinetic friction, not static friction, and I have [tex]sinθ[/tex] and [tex]cosθ[/tex], so that's two unknowns.

How do I find out the coefficient for static friction? Do I even need it?

And what went wrong with my analysis of this problem?

Thanks.
 

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leroyjenkens said:

Homework Statement



Here's a picture of the set up.

I need to find out what angle I need to have the plane raised to so that the blocks aren't moving.

Homework Equations



F = ma

The Attempt at a Solution


For the first block
[itex]F_k=μ_kF_N[/itex]

[itex]F_T+F_N-μ_kF_N+2mgcosθ+2mgsinθ=0[/itex]

Since the [itex]\ F_N\[/itex] and [itex]\ 2mg\cosθ\[/itex] cancel each other, I eliminated them from the problem. I set the right side equal to 0 since there is no acceleration.

Then I get [itex]\ F_T = μ_kF_N + 2mgsinθ[/itex]

For the right block [itex]\ mg = F_T[/itex]

So I input that into my equation for the other block and get [itex]\ mg=μ_kF_N+2mgsinθ[/itex]

For [itex]F_N[/itex] I plug in [itex]2mgcosθ[/itex]

And get [itex]\ mg=μ_k2mgcosθ+2mgsinθ[/itex]

And this is when I know something is wrong. That [itex]\ μ_k\[/itex] is the coefficient for kinetic friction, not static friction, and I have [itex]\ \sinθ\[/itex] and [itex]\ \cosθ\[/itex], so that's two unknowns.

How do I find out the coefficient for static friction? Do I even need it?

And what went wrong with my analysis of this problem?

Thanks.
For one thing, if the blocks are stationary, why not use μs, not μk ?

For another: [itex]\ F_N\[/itex] and [itex]\ 2mg\cosθ\[/itex] would not cancel each other in your expression. [itex]\ F_N\[/itex] is multiplied by μ .

The first thing to do is define a coordinate system for the 2m block and write equations for force for each component. It looks as though you have things all jumbled together into one (erroneous) equation.
 
Ok thanks.
 
Last edited:

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