# Finding g from Atwood's Machine

• jessicak
In summary, an Atwood's machine is a device used to measure the acceleration of gravity. When the masses are nearly equal, it is convenient to measure the acceleration and then calculate g. In this experiment, using masses of 450g and 452g, the masses moved a distance of 0.472m in 5.8s. This implies a value of g=13m/s/s, taking into account significant figures. However, there may have been experimental error in this setup.
jessicak
1.Homework Statement

Two masses hang from a string running over a pulley. Such a device can be used to measure the acceleration of gravity; it is then called Atwood's machine. If the masses are nearly equal, then the a of the masses will be much smaller than g; than makes it convienient to measure a and then calculate g. Suppose than an experimenter using masses m1= 450g and m2=452g finds that the masses move a distance of 0.472m in 5.8s starting from rest. What value of g does this imply? Assume the pulley is massless.

## Homework Equations

I know that a=(m2-m1/(m2+m1))g

## The Attempt at a Solution

From the above equation, I know that a=0.0022g
When using a=2y/t2, though I get g=12.8m/s/s, which is wrong (online hw).
Can anyone point me in the right direction?

jessicak said:
1.Homework Statement

Two masses hang from a string running over a pulley. Such a device can be used to measure the acceleration of gravity; it is then called Atwood's machine. If the masses are nearly equal, then the a of the masses will be much smaller than g; than makes it convienient to measure a and then calculate g. Suppose than an experimenter using masses m1= 450g and m2=452g finds that the masses move a distance of 0.472m in 5.8s starting from rest. What value of g does this imply? Assume the pulley is massless.

## Homework Equations

I know that a=(m2-m1/(m2+m1))g
yes, but you should derive that equation instead of looking it up.

## The Attempt at a Solution

From the above equation, I know that a=0.0022g
When using a=2y/t2, though I get g=12.8m/s/s, which is wrong (online hw).
Can anyone point me in the right direction?
I get g=12.6, which probably in either case should be rounded to 2 significant figures, g=13m/s/s.. Looks like there was a lot of experimental error in that setup.

## 1. What is Atwood's Machine?

Atwood's Machine is a device used to demonstrate the principles of acceleration and the relationship between mass and force.

## 2. How does Atwood's Machine work?

Atwood's Machine consists of two masses connected by a string that passes over a pulley. One side of the string has a heavier mass while the other side has a lighter mass. The force of gravity causes the heavier mass to accelerate downwards, while the lighter mass accelerates upwards at the same rate.

## 3. What is the purpose of finding g from Atwood's Machine?

Finding the value of g (the acceleration due to gravity) from Atwood's Machine allows for the calculation of other important physical quantities, such as the mass of the Earth and the weight of objects.

## 4. How do you calculate g from Atwood's Machine?

To calculate g, the following formula is used: g = (2m1m2)/(m1 + m2)a, where m1 and m2 are the masses on either side of the pulley and a is the acceleration of the system. This is derived from the equation F = ma, where F is the net force on the system and a is the acceleration.

## 5. What are some sources of error when finding g from Atwood's Machine?

Some sources of error include friction in the pulley and string, air resistance, and imprecise measurements of mass and acceleration. These factors can cause the calculated value of g to differ slightly from the actual value.

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