# Half Atwood machine: acceleration after inital force

• arturo
In summary, the conversation discusses the acceleration of a hanging block after the block on the table is given a push to the right or left. The solution to part (a) is correct, but the solution to part (b) is different due to the direction of frictional force changing. The short time interval after the force is removed is still impacted by the initial velocity in the opposite direction, leading to a different final acceleration.
arturo

## Homework Statement

Assume that the block on the table (Figure 1) has twice the inertia of the hanging block.

(a)You give the block on the table a push to the right so that it starts to move. If the magnitude of the frictional force exerted by the table on the block is half the magnitude of the gravitational force exerted on the hanging block, what is the acceleration of the hanging block after you have stopped pushing the block on the table?

(b)What would this hanging-block acceleration be if you had pushed the block on the table to the left instead? Consider only a short time interval after you stop pushing.

Fnet = ma
Fg = mg

## The Attempt at a Solution

I did get an answer to part(a). I solved as follows
Ftable = T - .5mg = 2ma
Fmass = T - mg = -ma
(set equations equal to T)
-ma + mg = 2ma + .5mg
-a + g = 2a + .5g
3a = .5g
a = 1/6g

And that was correct. I assumed the initial force to get the system moving had no part to play in the equations.
When I looked at part(b), I thought it would be the exact same, the forces in my equations had not changed even though there was a initial force in the -x direction that was no longer present. However, the answer is not the same, and I don't understand how that momentary initial force impacts the final acceleration.

arturo

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arturo said:
When I looked at part(b), I thought it would be the exact same, the forces in my equations had not changed
Does each force in part (b) still have the same direction as in part (a)?

arturo
Friction changes signs when the force is applied in the opposite direction on the table block, and I would be lead to assume that that stays facing the opposite direction even after the force is no longer applied.

If that’s the case, then:
-a + g = 2a - .5g
3a = 1.5g
a = .5g

Which is correct.

For a moment I did not understand why the frictional force wouldn’t flip back the moment the force on the table block was no loner exerted. After reconsidering the problem, that would be because within that short time interval after the force is removed, it would still have velocity in the -x direction (as friction opposes motion, not force), correct?

arturo said:
For a moment I did not understand why the frictional force wouldn’t flip back the moment the force on the table block was no loner exerted. After reconsidering the problem, that would be because within that short time interval after the force is removed, it would still have velocity in the -x direction (as friction opposes motion, not force), correct?
Yes. Good work.

## 1. What is a Half Atwood machine?

A Half Atwood machine is a simple mechanical device that consists of a pulley, a mass attached to one end of a string, and a support that the other end of the string is attached to. It is used to study the effects of forces and acceleration in a controlled environment.

## 2. How does the Half Atwood machine work?

The Half Atwood machine works by applying a force on one end of the string, which causes the mass to accelerate due to the force of gravity. The pulley allows the string to move freely, reducing friction and allowing for more accurate measurements.

## 3. What is the acceleration of the mass in a Half Atwood machine?

The acceleration of the mass in a Half Atwood machine can be calculated using Newton's Second Law, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the difference between the force applied and the force of gravity, and the mass is the weight of the object.

## 4. How does the initial force affect the acceleration in a Half Atwood machine?

The initial force applied to the Half Atwood machine will determine the acceleration of the mass. The greater the initial force, the greater the acceleration will be. This is because the initial force adds to the force of gravity, resulting in a larger net force and therefore a larger acceleration.

## 5. What factors can affect the acceleration in a Half Atwood machine?

Aside from the initial force, the acceleration in a Half Atwood machine can also be affected by the mass of the object, the angle of the string, and any external forces such as friction. These factors can influence the net force acting on the object and therefore affect its acceleration.

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