Friction - 3 Blocks and a Pulley

• FredericChopin
In summary: Thank you for the feedback! In summary, the force of static friction on Block 1 due to Block 2 can be calculated using Newton's Second Law, which states that the sum of all forces in the x-direction is equal to the mass of the object times its acceleration. The force of static friction can take on any value up to a maximum value of μ*m1*g, and it is the only force acting in the x-direction on Block 1.
FredericChopin

Homework Statement

http://imgur.com/C0XYEKw

Fnet = m*a
FG = m*g
fstatic = μ*N

The Attempt at a Solution

If both Block 1 and Block 2 have acceleration a in the x-direction and have an acceleration of 0 in the y-direction, then the force of static friction on Block 1 due to Block 2 doesn't seem to depend at all on what Block 2 is doing. This is because the normal force on Block 1 due to Block 2 is still the same as it is at rest and the force of static friction on Block 1 due to Block 2 is the only force acting in the x-direction on Block 1.

I tried this by considering only Block 1 as a system.

In the y-direction, the forces acting on Block 1 are the normal force on Block 1 due to Block 2, N12, and the gravitational force on Block 1 due to the Earth, F1E. The acceleration, ay, is equal to 0.

Thus:

Σ Fnet,y = 0 = N12 - F1E = N12 - m1*g

Thus:

N12 = m1*g

In the x-direction, the only force acting on Block 1 is the force of static friction, which is acting in the right, positive x-direction. The acceleration, ax, is equal to a.

Thus:

Σ Fnet, x = m1*a = fstatic = μ*N12 = μ*m1*g

Thus the force and direction of the force of static friction on Block 1 due to Block 2 is:

fstatic = μ*m1*g

This, however, is the wrong answer.

I don't understand why though. Where am I going wrong?

Thank you.

FredericChopin said:

Homework Statement

http://imgur.com/C0XYEKw

Fnet = m*a
FG = m*g
fstatic = μ*N

The Attempt at a Solution

If both Block 1 and Block 2 have acceleration a in the x-direction and have an acceleration of 0 in the y-direction, then the force of static friction on Block 1 due to Block 2 doesn't seem to depend at all on what Block 2 is doing. This is because the normal force on Block 1 due to Block 2 is still the same as it is at rest and the force of static friction on Block 1 due to Block 2 is the only force acting in the x-direction on Block 1.

I tried this by considering only Block 1 as a system.

In the y-direction, the forces acting on Block 1 are the normal force on Block 1 due to Block 2, N12, and the gravitational force on Block 1 due to the Earth, F1E. The acceleration, ay, is equal to 0.

Thus:

Σ Fnet,y = 0 = N12 - F1E = N12 - m1*g

Thus:

N12 = m1*g

In the x-direction, the only force acting on Block 1 is the force of static friction, which is acting in the right, positive x-direction. The acceleration, ax, is equal to a.

Thus:

Σ Fnet, x = m1*a = fstatic = μ*N12 = μ*m1*g

Thus the force and direction of the force of static friction on Block 1 due to Block 2 is:

fstatic = μ*m1*g

This, however, is the wrong answer.

I don't understand why though. Where am I going wrong?

Thank you.

What is the difference between kinetic and static friction?
Imagine that you have a block in rest on a table. The mass of the block is 2 kg, the coefficient of static friction is μs=0.5.

You push the block with force F=1N force. Will the block move? What is the force of friction?
You push the block with F=9 N force. Will the block move? What is the force of friction now?

O.K., I realized I grossly overcomplicated things.

Since the force of static friction on Block 1 due to Block 2 is the only force in the x-direction acting on Block 1, and Block 1 has acceleration a, then Newton's Second Law states that:

∑ Fx = m1*a = fs12

Thus, the force of static friction on Block 1 due to Block 2 is:

fs12 = m1*a

EDIT: Responding to your original question, the force of static friction can take on any value up to a maximum value of μ*m1*g. So, when an applied force of 1 N is acting on Block 1, the force of static friction is 1 N in the opposite direction. As a result, the net force is 0 N and Block 1 does not move. If an applied force of 9 N is acting on Block 1, the force of static friction is at its maximum 9 N (if g = 10 m/s2) in the opposite direction. As a result, Block 1 travels at a constant velocity in the direction of applied force.

Last edited:
FredericChopin said:
O.K., I realized I grossly overcomplicated things.

Since the force of static friction on Block 1 due to Block 2 is the only force in the x-direction acting on Block 1, and Block 1 has acceleration a, then Newton's Second Law states that:

∑ Fx = m1*a = fs12

Thus, the force of static friction on Block 1 due to Block 2 is:

fs12 = m1*a
That is correct now.

1. What is friction?

Friction is the force that resists the motion of an object when it comes into contact with another surface.

2. How does friction affect the movement of 3 blocks connected by a pulley?

Friction can impact the movement of 3 blocks connected by a pulley by creating resistance and reducing the efficiency of the system. This means that more energy is needed to move the blocks and the pulley may not work as smoothly.

3. How can friction be reduced in this scenario?

Friction can be reduced by using lubricants, such as oil, between the surfaces of the blocks and the pulley. Additionally, using smoother and more polished surfaces can also help decrease friction.

4. Can friction be completely eliminated?

No, friction cannot be completely eliminated. It is a natural force that exists between any two surfaces that come into contact with each other.

5. How does the weight of the blocks affect friction in this system?

The weight of the blocks can increase the amount of friction in the system, as there is more force pressing the surfaces together. This can make it more difficult to move the blocks and the pulley.

Replies
4
Views
2K
Replies
6
Views
395
Replies
61
Views
1K
Replies
15
Views
384
Replies
17
Views
2K
Replies
16
Views
2K
Replies
1
Views
1K
Replies
23
Views
2K
Replies
33
Views
581
Replies
23
Views
2K