# Angle required to keep block from sliding down plane

## 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.

F = ma

## The Attempt at a Solution

For the first block
$$F_k=μ_kF_N$$
$$F_T+F_N-μ_kF_N+2mgcosθ+2mgsinθ=0$$

Since the $$F_N$$ and $$2mgcosθ$$ 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
$$F_T = μ_kF_N + 2mgsinθ$$

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

For $$F_N$$ I plug in $$2mgcosθ$$

And get $$mg=μ_k2mgcosθ+2mgsinθ$$

And this is when I know something is wrong. That $$μ_k$$ is the coefficient for kinetic friction, not static friction, and I have $$sinθ$$ and $$cosθ$$, 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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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.

F = ma

## The Attempt at a Solution

For the first block
$F_k=μ_kF_N$

$F_T+F_N-μ_kF_N+2mgcosθ+2mgsinθ=0$

Since the $\ F_N\$ and $\ 2mg\cosθ\$ 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 $\ F_T = μ_kF_N + 2mgsinθ$

For the right block $\ mg = F_T$

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

For $F_N$ I plug in $2mgcosθ$

And get $\ mg=μ_k2mgcosθ+2mgsinθ$

And this is when I know something is wrong. That $\ μ_k\$ is the coefficient for kinetic friction, not static friction, and I have $\ \sinθ\$ and $\ \cosθ\$, 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: $\ F_N\$ and $\ 2mg\cosθ\$ would not cancel each other in your expression. $\ F_N\$ 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.

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