Solve 2-Mass Pulley System Homework for Moving Blocks

[dgZeyas]

Homework Statement

A massless string, passing over the frictionless pulley, connects the two masses as shown. Block m₁ hangs vertically without touching the wall, but m₂ slides along the wall with friction (μ_s = 0.61, and μ_k = 0.14). If θ = 53°, how many times heavier than block 1 must block 2 be to start the system moving?

Homework Equations

1. ƩF_x = ma_x and ƩF_y = ma_y

2. f_s = μ_sN
3. f_k = μ_kN

The Attempt at a Solution

I set up a free body diagram for m₂, oriented as it is in space, such that the positive x-direction points up the slope and the positive y-direction points perpendicularly out of the slope. Then the forces are as follows:

T (tension): + x-direction
W (weight): m₂gsinθ in the -x-direction and m₂gcosθ in the -y-direction
N (normal): +y direction
f_s (static friction): +x-direction (when m₂ is NOT moving)
f_k (static friction): -x-direction (when m₂ IS moving)

So I figured the static friction would be this:

f_s= μ_s(m₂)(g)(sinθ)

as per equation 2.

But this is about as far as I got. I'm trying to find how many times heavier than block 1 must block 2 be to start the system moving-- I inferred from this that block 2 would move UP the ramp if block 1 was heavier, which makes sense, and while I have an equation containing m₂, I'm having trouble figuring out another equation so a ratio can be set up.

Any help appreciated, thanks in advance!

 

[collinsmark]

So I figured the static friction would be this:

f_s= μ_s(m₂)(g)(sinθ)

as per equation 2.

(Emphasis in red, mine.)

Are you sure about that? (Hint: what's the normal force on m₂?)

After that, the next step is to ask yourself, "if nothing is moving, what is the tension in the string?"

 

[dgZeyas]

Oh, yeah, ok it'd be cosθ. Because the normal force is the positive y-component of the weight. Thanks for pointing that out.

Tension (according to my coordinate system) is in the positive x-direction. OK. So if nothing is moving, ƩF_x=0, as per equation 2 and Newton's second law.

So ƩF_x = T + f - Wsinθ = 0, where W is the weight of the second mass.

So T = Wsinθ/f. f is the static friction ,or μ_s multiplied by the normal force... which is a vector in the y-direction, not the x, so now I confused how to add that bit in.

Thanks for the pointers so far.

Edit: I got a bit further and I can calculate the tension in the string using a FBD for mass 1. So that would enable me to find the static friction force, but I don't particularly see how that gets me closer to the answer either to be honest.

 

[collinsmark]

Oh, yeah, ok it'd be cosθ. Because the normal force is the positive y-component of the weight. Thanks for pointing that out.

Tension (according to my coordinate system) is in the positive x-direction. OK. So if nothing is moving, ƩF_x=0, as per equation 2 and Newton's second law.

So ƩF_x = T + f - Wsinθ = 0, where W is the weight of the second mass.

So T = Wsinθ/f. f is the static friction ,or μ_s multiplied by the normal force... which is a vector in the y-direction, not the x, so now I confused how to add that bit in.

Thanks for the pointers so far.

Edit: I got a bit further and I can calculate the tension in the string using a FBD for mass 1. So that would enable me to find the static friction force, but I don't particularly see how that gets me closer to the answer either to be honest.

You're doing great!

But don't forget! You've already found an expression for the static friction force in your first post (with the cos correction in your last post),
f_s = μ_sm₂g cosθ.
So feel free to substitute that in.

You've also mentioned that by using mass m₁ that you can also find another expression with the tension in it, so that get's rid of the T variable.

Now just express m₂/m₁ as a function of θ and μ_s, and that's pretty much it.

 

[collinsmark]

Oh, and welcome to Physics Forums, by the way!

 

[dgZeyas]

Long story short-- I've gotten it all figured out. As my professor says, non-calc physics is just a LOT of algebra. Well, yes it is, yes it is.

Anyways, thanks very much for all your help and your welcome!

Feel free to lock or delete the thread or whatever :)

Thanks again :)