Angle required to keep block from sliding down plane

[leroyjenkens]

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
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.

 

[SammyS]

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
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 μ_s, not μ_k ?

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.

 

[leroyjenkens]

Ok thanks.