Find Acceleration & Tension in a System with Two Blocks and a Connecting Cord

[sunnnystrong]

Homework Statement

Determine the acceleration of the system and the tension in the connecting cord.

Mass 1 = 2.8kg
T1 = ?
Mass 2 = 1.2kg
T2 = ?
mu = .2

*My picture is kinda bad but you can see how its set up*

Homework Equations

∑F = m*a

The Attempt at a Solution

I drew a FBD for Block A -->
∑Fx = m*a

T1 - ƒ = m*a
T1 = m*g*mu + m*a

For Block B -->
Down = positive
T2 = mg - ma My question is... are the two tensions equal? Would I just set these two equations equal to each other & solve for a and than solve for T?

 

[gneill]

Assuming that there is light, frictionless pulley where the cord makes a right angle turn then yes, the tension will be the same everywhere in the cord.

 

[sunnnystrong]

Assuming that there is light, frictionless pulley where the cord makes a right angle turn then yes, the tension will be the same everywhere in the cord.

So I guess m cancels?
For a I got :
a = 1/2(g-mu*g)
a = 3.92 m/s/s

Than: For block 2 (mass = 1.2kg)
T2 = mg - ma
T2 = 7.056 N

For block 1 (mass = 2.8kg)
T1 = m*g*mu + m*a
T1 = 16.464 N

So what tension is it?

 

[TomHart]

So I guess m cancels?
For a I got :
a = 1/2(g-mu*g)

Can you explain where this equation came from?

And if it is a frictionless pulley, then T1 and T2 should be equal, true?

 

[sunnnystrong]

Can you explain where this equation came from?

And if it is a frictionless pulley, then T1 and T2 should be equal, true?

T1 = m*g*mu + m*a
T2 = mg - ma
m*g*mu + m*a = mg - ma
g*mu + a = g - a
2a = g - g*mu
a = (1/2)(g - g*mu)

I'm kinda confused?

 

[TomHart]

You can't cancel out the masses because they aren't all the same. Some are m1 and some are m2.

 

[sunnnystrong]

You can't cancel out the masses because they aren't all the same. Some are m1 and some are m2.

Oh wow hahahah thank you
XD