Finding the coefficient of kinetic friction of 3 blocks

[chevymechanic]

Homework Statement

When three blocks are released from rest, they accelerate with a magnitude of 0.500 m/s². 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?
(The blocks are arranged with block 1 hanging from the left side of the table, block 2 sitting on the middle of the table, and block 3 hanging from the right side of the table. All connected by an ideal rope.)

Homework Equations

f_k=u_kF_N
F=ma

The Attempt at a Solution

This is the problem. I have no clue where to start. The only thing I know is that the coefficient of friction is u_k and has no units. I understand that I must use the first formula somehow. Would I first find the normal force of block 2? I'm lost...

 

[PeterO]

Homework Statement

When three blocks are released from rest, they accelerate with a magnitude of 0.500 m/s². 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?
(The blocks are arranged with block 1 hanging from the left side of the table, block 2 sitting on the middle of the table, and block 3 hanging from the right side of the table. All connected by an ideal rope.)

Homework Equations

f_k=u_kF_N
F=ma

The Attempt at a Solution

This is the problem. I have no clue where to start. The only thing I know is that the coefficient of friction is u_k and has no units. I understand that I must use the first formula somehow. Would I first find the normal force of block 2? I'm lost...

Think globally.

The total system has mass 6M

A force equal to the weight of 2M is trying to pull it one way, while a force equal to the weight of 3M tries to pull it the other way.

That means a net force equal to the weight of 1M is accelerating the masses in a given direction.

The "weight of 1M" acting on a total of 6M should give an acceleration you can calculate.

However, due to friction between the 1M block and the table, the acceleration is reduced to only 0.5. from that you can calculate the size of the firction force, and thus the coefficient of friction.

 

[chevymechanic]

Wouldn't the total mass be 5M? The block hanging from the left side of the table is M, middle block is 2M, and the block hanging from the right side is also 2M. I guess it would still mean that a mass of 1M is accelerating the masses though. I don't understand why I would have to calculate the acceleration when it is given as .500 m/s² in the problem. Would I use F=ma?

 

[PeterO]

Wouldn't the total mass be 5M? The block hanging from the left side of the table is M, middle block is 2M, and the block hanging from the right side is also 2M. I guess it would still mean that a mass of 1M is accelerating the masses though. I don't understand why I would have to calculate the acceleration when it is given as .500 m/s² in the problem. Would I use F=ma?

You're right. Total 5M. Mis-read and with no diagram.

If you calculate what the acceleration "would have been" in the absence of friction, you can easily assess the effect of friction - how strong it is.

You can calculate via forces, but I find it easier to work through acceleration.

eg with g=10

Net force equivalent to on 1M, but system mass 5M thus acceleration only 1/5 th of normal [without friction] so 2 m/s².

Actual acceleration only 0.5, so 3/4 of the net force above is "cancelled by friction" Thus Friction is 3/4 the weight force of M - which is 3/8 or 2M (the Normal Reaction Force for the mass on the table).

Now when you use g = 9.8 or 9.81 you get decimals all over the place, but the idea is the same.

What a pity the standard metre wasn't just a little bit shorter, so that g actually equalled 10.

 

[asteriatic]

I would first start by defining the variables and parameters in the problem. In this case, we have three blocks with masses M, 2M, and 2M, and they are connected by an ideal rope. The blocks are arranged in a specific way, with block 1 hanging from the left side of the table, block 2 sitting on the middle of the table, and block 3 hanging from the right side of the table. The acceleration of the blocks is given as 0.500 m/s2.

Next, I would apply the relevant equations to the problem. From Newton's second law (F=ma), we can calculate the net force acting on each block. Since the blocks are accelerating, there must be a net force acting on them. We can also use the equation for kinetic friction (fk=ukFN) to relate the frictional force to the normal force and the coefficient of kinetic friction.

In this case, we are interested in finding the coefficient of kinetic friction between block 2 and the table. To do this, we need to find the normal force acting on block 2. This can be done by considering the forces acting on the block in the vertical direction. Since the block is sitting on the table, the normal force must be equal in magnitude to the weight of the block (mg), where g is the acceleration due to gravity.

Once we have the normal force, we can use the equation for kinetic friction to find the frictional force acting on block 2. Since we know the acceleration of the blocks (0.500 m/s2), we can use Newton's second law to find the net force acting on block 2. This net force must be equal to the sum of the frictional force and the weight of the block. From this, we can solve for the coefficient of kinetic friction (uk) between block 2 and the table.

In summary, to find the coefficient of kinetic friction between block 2 and the table, we need to use the equations for Newton's second law and kinetic friction, and consider the forces acting on block 2 in the vertical direction. By solving for the unknown variables, we can find the coefficient of kinetic friction and complete the problem.