The aim of the unbalanced forces experiment is to study and analyze how an unbalanced forces effects an object. To achieve this objective, the Atwood Machine is used to where a set of pennies are placed in either masses of the machine. The Atwood Machines uses the principle of the Newton’s 2nd law of motion which attributes the force on a moving object to be the product of the objects acceleration and it’s mass. In the case of an Atwood machine, the string containing the two mass has a tension, T, which is constant. Therefore, each mass on the Atwood Machine is acted upon by:

〖F-F〗_T=ma

For the smaller mass, the equation becomes;

〖F_m-F〗_T=ma

While for the larger mass, the equation becomes;

〖F_M-F〗_T=Ma

Taking F=mg and Mg, then the acceleration can be found to be;

a=((M-m))/(M+m) g

The acceleration is calculated using the formula above and then compared to the acceleration measured from the linear fit of velocity-time graph to compute the % error.

Error Analysis

There were five trials in the experiment arising due to varying of masses in both sides of the Atwood Machine. The average % error of the acceleration from the five trials was:

=(19.28%+20.63%+16.07%+12.25%+15.28%)/5=16.702%

This % error alongside the fluctuations seen in the standard deviation of the obtained acceleration indicate that systematic errors were present in the experiment. The systematic errors are caused due to the friction of the Atwood Machine and uncertainties in the measurement of the mass of pennies. The error can only be reduced by lubricating the Atwood machine to reduce friction and by taking several mass measurements then finding the average to reduce the uncertainty.

Expansion Questions

When the smaller mass m is zero, the acceleration equation becomes:

a=(M/M)g

a=g

This makes sense in case a where on mass M falls freely in space the acceleration a and g will be equal and towards the same direction. When the larger mass M is zero, the acceleration equation becomes:

a=(m/m)g

a=-g

This does not make sense since when mass m falls freely the acceleration a and g will be equal and towards the same direction and not vice versa.