Friction - 3 Blocks and a Pulley

[FredericChopin]

Homework Statement

http://imgur.com/C0XYEKw

Homework Equations

F_net = m*a
F_G = m*g
f_static = 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, F_1E. The acceleration, a_y, is equal to 0.

Thus:

Σ F_net,y = 0 = N₁₂ - F_1E = 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, a_x, is equal to a.

Thus:

Σ F_{net, x} = m₁*a = f_static = μ*N₁₂ = μ*m₁*g

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

f_static = μ*m₁*g

This, however, is the wrong answer.

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

Thank you.

 

[ehild]

Homework Statement

http://imgur.com/C0XYEKw

Homework Equations

F_net = m*a
F_G = m*g
f_static = 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, F_1E. The acceleration, a_y, is equal to 0.

Thus:

Σ F_net,y = 0 = N₁₂ - F_1E = 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, a_x, is equal to a.

Thus:

Σ F_{net, x} = m₁*a = f_static = μ*N₁₂ = μ*m₁*g

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

f_static = 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 μ_s=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?

 

[FredericChopin]

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:

∑ F_x = m₁*a = f^s₁₂

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

f^s₁₂ = 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.

 

[ehild]

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:

∑ F_x = m₁*a = f^s₁₂

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

f^s₁₂ = m₁*a

That is correct now.