Kinetic Friction and Ropes and Pulleys

[dizkoducky]

1.
Homework Statement
Block A, on a horizontal table, has a weight of 4.91N is attached to a rope, that goes over a pulley, attached to Block B, that has a weight of 2.94N. Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible. Need to find the kinetic friction coeffecient for Block A and Block B.

Homework Equations

Kinetic Friction = (coeffecient of Kf)(Normal Force) in the opposite direction
Tension - Weight = 0
Therefore, Tension = Weight

The Attempt at a Solution

I have drawn FBDs for both blocks. I understand the tension & normal force of the two, individually. I am having trouble connecting the tension to the normal force to find the coeffecient.

Thanks!
starr

 

[MATdaveLACK]

I don't understand why you need to find the coefficient of friction on Block B unless Block B is sliding along something as it falls.

As for your question, let's look at Block A. Along the horizontal axis you have Tension pulling the block and friction resisting the block's motion. These two forces will be equal since the blocks are moving at a constant speed. In the vertical axis you have the normal force pushing up on Block A and Fg pushing down. Again these two forces will be equal.
So we have:
(for Block A)
Tension = Ff
Normal force = Fg

Once you have Normal force and Ff you can solve for the coefficient of Friction.But like I said, if there is friction acting on block B the problem will be a little different. (Specifically the Tension will change)

 

[baba89]

ider

Hello starrider,

In this scenario, the tension in the rope is equal to the weight of the blocks (Tension = Weight). This means that there is no net force acting on the system, and the blocks are in equilibrium.

Since the blocks are moving at a constant speed, this means that the frictional force acting on them must be equal and opposite to the tension force. We can use this information to find the coefficient of kinetic friction for both blocks.

First, let's focus on Block A. The frictional force acting on Block A is equal to the coefficient of kinetic friction (μ) multiplied by the normal force (N). Since the block is on a horizontal table, the normal force is equal to the weight of the block (N = mg). Therefore, we can write the equation:

Ff = μmg

Since we know the weight of Block A (4.91N) and the coefficient of kinetic friction is what we are trying to find, we can rearrange the equation to solve for μ:

μ = Ff/mg

Now, let's look at Block B. The frictional force acting on Block B is also equal to the coefficient of kinetic friction (μ) multiplied by the normal force (N). In this case, the normal force is equal to the weight of Block B plus the tension force (N = mg + T). Since we know that T = mg, we can write the equation:

Ff = μ(mg + mg)

Again, we can rearrange the equation to solve for μ:

μ = Ff/2mg

Now, we can plug in the values for the frictional force for both blocks. Since the blocks are moving at a constant speed, the frictional force must be equal to the tension force, which is equal to the weight of the blocks. Therefore, we can write the equations:

For Block A: μ = 4.91N/4.91N = 1
For Block B: μ = 2.94N/2(4.91N) = 0.3

So, the coefficient of kinetic friction for Block A is 1 and the coefficient of kinetic friction for Block B is 0.3.

I hope this helps! Let me know if you have any other questions. Keep up the good work in your studies of kinetic friction, ropes, and pulleys!