Angle required to keep block from sliding down plane

  • #1
575
47

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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  • #2
SammyS
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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
[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.
 
  • #3
575
47
Ok thanks.
 
Last edited:

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