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## Homework Statement

When the three blocks in Fig. 6-33 are released from rest, they accelerate with a magnitude of 0.800 m/s^2. Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coefficient of kinetic friction between block 2 and the table?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/pict_6_33.gif

## Homework Equations

F_f = mu_k*F_N

## The Attempt at a Solution

So far I have:

The x-component of the force on the second block is equal to the sum of the y-component of 1, the y-component of 3, and the friction between the table and 2 :

F_2,x = 2*M(0.8) = F_1,y + F_3,y + F_f

For friction, we have:

F_f = mu_k*F_N = mu_k*2*M*g

Since the normal force is just gravity in this case. Substituting, we have

F_2,x = 2*M*g - (M*g + mu_k*2*M*g) = (1 - 2*mu_k)*M*g

Since the acceleration is 0.8 then we solve for mu_k

F_2,x = (0.8)*(2.0)*M = 1.6*M = (1 - 2*mu_k)*M*g

1.6 = (9.8)*(1 - 2*mu_k)

(-1/2)*((1.6/9.8) - 1) = mu_k

(-1/2)*(-8.2/9.8) = 0.418

But this answer isn't right. I have a hunch I messed up the beginning by assuming that

F_2,x = 2*M(0.8) = F_1,y + F_3,y + F_f

But I'm not sure exactly.