Forces help (friction, tension, etc.)

1. Mar 3, 2015

Reizor1989

First off, I want to apologize if anything is made unclear. I'm ultra-stressed right now, I'm sure I can get over this physics problem. I just need a layman's explanation of this.

1. A taut massless string connects two boxes as shown below. The boxes are placed on an incline plane at θ = 33.00. What is the acceleration of boxes as they move down the incline, given m2 = 1.2 kg, m1 = 6.8 kg, μ2 = 0.10 and μ1 = 0.20 .

2. Constructed free body diagram for M1 (mass 1) with normal force (n) along the positive y-axis, friction (f) along the negative x-axis, tension (T) along the positive x-axis, and Force of gravity (Fg) in quadrant IV, 33 degrees from the negative y-axis. (a=acceleration, mu=coefficient of friction)

a.) Summed forces of x as Fx = T - f + Fgx = M1*a ->
-> M1 * a - mu*n + M1*g = M1 * a
-> (This is where I think I'm screwing up. I want to isolate the first (a), but I have no idea how.)

b.) Summed forces of y as Fy = n - Fgy = M1 * a ->
-> n - Fgy = 0 (acceleration is 0)
-> n = Fgy

3. Constructed free body diagram for M2 (mass 2) with (n) along the positive y-axis, (f) along the negative x-axis, tension (T) along the positive x-axis, and Force of gravity (Fg) in quadrant IV, 33 degrees from the negative y-axis. (a=acceleration, mu=coefficient of friction)

a.) Same as 2a
b.) Same as 2b

4. Basically I have no idea where to go from here. Did I screw up on the diagram? Is something wrong with my algebra? What can I do next?

2. Mar 3, 2015

Staff: Mentor

Welcome to the PF.

It sounds like you are setting up your x-y coordinates so that x is parallel to the inclined plane. It is more traditional to have x be horizontal, parallel to the ground, with y vertical. Then you write the forces on the inclined plane for the FBDs in terms of sin(θ) and cos(θ) components.

Can you re-write your FBD equations with the x-y axes in the more traditional orientation? That will make it a lot easier to check your work.

3. Mar 3, 2015

TSny

Welcome to PF!

Look's like you assumed T = M1a. Only the net force on M1 equals M1a. For Fgx it looks like you didn't get the x component correct.

Good. You'll just need to get an explicit expression for Fgy.

What direction is the tension on M2?

Treating T and $a$ as unknowns you should be able to solve for $a$.

[EDIT: I see I wasn't quick enough. berkeman posted while I was still constructing my response. I think it's good to choose the orientation of the axes as you did. Then the acceleration has only an x component.]