Question

When the three blocks in the figure below are released from rest, they accelerate with a magnitude of 0.705 m/s2. Block 1 has mass M, block 2 has 4M, and block 3 has 4M. What is the coefficient of kinetic friction between block 2 and the table?

Answer 1

You're missing the figure.

Answer 2

Can you elaborate on what you mean by the figure?

Answer 3

Oh yes I will attach it momentarily

Answer 4

are the pulley's frictionless?

Answer 5

Yes they are frictionless

Answer 6

The first thing would be to draw a FBD.

Answer 7

Okay I have the FBD drawn for each mass block

Answer 8

Ok I denoted mine using the numbers so hopefully this doesn't confuse you.
\(T_1-m_1g=m_1a\)
\(F_2-f_k=0\)
\(m_3g-T_3=m_3a\)

Answer 9

Okay so the first thing I've done is plugged in my values for the first equation to find T1, am I on the right track? If so, what is my next step?

Answer 10

Unfortunately, it's dinner time.
I'm rusty on mechanics anyways so I tend to forget stuff.
@douglaswinslowcooper

Answer 11

Ja ne! \(\huge ☺\)

Answer 12

Thank you!

Answer 13

Accelerating force = net weight times gravity - friction

F = (4 M - M)g - 4 M mu mu = kinetic friction coefficient

mass being accelerated = M + 4M + 4M

a = 0.705 = F/mass = [(4 M - M)g - 4 M mu]/(M + 4M + 4M)

a = 0.705 = (3 g-4 mu)/(9) solve for mu

g= acceleration of gravity 9.8m/s^2

Answer 14

Thank you, my last question in regard to this is if I am supposed to multiply this value 3g to be (3*9.8)? I really appreciate the help

Answer 15

yes, in mks units 3 g = 3 x 9.8 m/s^2
Glad to have helped.