Finding static friction of a block on a table on a pulley?

[Eclair_de_XII]

Homework Statement

"When the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of $0.500 \frac{m}{s^2}$. Block 1 has mass M, Block 2 has mass 2M, and Block 3 has mass 3M. What is the coefficient of static friction between Block 2 and the table?"

Homework Equations

$f_s=(μ_k)(F_N)$
Answer from book: 0.37

The Attempt at a Solution

Okay, so I attempted this by beginning with the sum of all the forces at work, relative to the force on Block 3. There's the gravitational force on Block 1 negating half of Block 3's force, and then there's the static friction between Block 2 and the table also slowing the motion of Block 3. So:

$ΣF = ma = (-9.8\frac{m}{s^2})(M) + (μ_k)(-9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(2M)$
$ΣF = (μ_k)(-9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(M)$
$\frac{ΣF}{m} = a = (9.8\frac{m}{s^2})(-2μ_k + 1)$
$a = \frac{1}{2}\frac{m}{s^2}$
$\frac{1}{2}\frac{m}{s^2} = (9.8\frac{m}{s^2})(-2μ_k + 1)$
$\frac{1}{19.6}=-2μ_k + 1$
$-\frac{18.6}{19.6}=-2μ_k$
$μ_k=\frac{93}{196}$

I have a feeling I messed up on the third step, in calculating acceleration and equating it to $\frac{1}{2}$. Are there any other forces at work here, that are slowing Block 3's descent? Thank you for anyone who is willing to help me.

 

[gneill]

Your problem statement gives block three a mass of 3M, but it looks like you've used 2M in your first step. Can you verify which is correct?

What mass are you using for 'm' in F = ma? Hint: How much mass is accelerating?

 

[Eclair_de_XII]

Can you verify which is correct?

Right. I should be more mindful of reading the problem carefully; especially when doing physics homework.

$ΣF = -(μ_k)(9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(2M)$

How much mass is accelerating?

6M?

$\frac{ΣF}{6M} = -\frac{1}{3}(μ_k)(9.8\frac{m}{s^2}) + \frac{1}{3}(9.8\frac{m}{s^2})$
$\frac{ΣF}{6M} = \frac{1}{3}(9.8\frac{m}{s^2})(1-μ_k) = a$
$\frac{1}{2} = \frac{1}{3}(9.8\frac{m}{s^2})(1-μ_k)$
$\frac{3}{19.6} = 1 - μ_k$
$-μ_k=-\frac{16.6}{19.6}$
$μ_k=\frac{16.6}{19.6}$

I wish I knew what I was doing wrong.

 

[gneill]

You say the total amount of mass in the problem is 6M, so then the blocks have masses 1M, 2M, and 3M.

When I do the math assuming those block masses I don't get the book answer. On the other hand, If the third block is 2M and the total mass is 5M, then I get the given answer.

You might want to use a symbol for the gravitational acceleration (g) rather than lug around numbers and units all the way through the algebra. Only plug in numbers at the very end.

 

[Eclair_de_XII]

On the other hand, If the third block is 2M and the total mass is 5M, then I get the given answer.

Oops, I should learn to read more carefully.

$ΣF = (-μ_k)(g)(2M) + (g)(M)$
$a = \frac{1}{5}g(1-2μ_k)$
$\frac{1}{2} = \frac{1}{5}g(1-2μ_k)$
$\frac{5}{2g} = 1-2μ_k$
$\frac{5-2g}{2g} = -2μ_k$
$\frac{2g-5}{4g}=μ_k=0.37$

Only plug in numbers at the very end.

Good to note. Thank you...