The Atwood MachineThe Atwood Machine

Formal Lab ReportFormal Lab Report

Atwood’s MachineAtwood’s Machine The purpose of this The purpose of this

laboratory exercise is to laboratory exercise is to investigate the validity investigate the validity of Newton’s Second Law of Newton’s Second Law of Motion. You will use a of Motion. You will use a single pulley apparatus single pulley apparatus as shown. This as shown. This apparatus is commonly apparatus is commonly referred to as an referred to as an “Atwood Machine”. The “Atwood Machine”. The Atwood Machine allows Atwood Machine allows one object to hoist one object to hoist another object using another object using only gravity.only gravity.

40 cm

Atwood’s MachineAtwood’s Machine

To determine the tension in the To determine the tension in the string (Fstring (FTT), we have to draw a free-), we have to draw a free-body diagrambody diagram

M1M2

Fgm1 Fgm2

FT FT

Making Sense Of It!Making Sense Of It!

FFm2m2 = m = m22a = ma = m22 g – F g – FTT Eqn 1Eqn 1

andand FFm1m1 = m = m11a = -ma = -m11 g + F g + FTT Eqn 2Eqn 2

How can we simplify this?How can we simplify this?

Here’s How!Here’s How!

We can determine the FWe can determine the Fgg’s because ’s because we know the masses involved and g we know the masses involved and g is always 9.8 m/sis always 9.8 m/s22..

The problem is FThe problem is FT T we don’t know we don’t know what it is and we can not calculate what it is and we can not calculate it!?!it!?!

Lets get rid of it! Lets get rid of it! Make a Make a substitutionsubstitution

Substitute for FSubstitute for FTT

If If mm22a = ma = m22 g – F g – FTT, ,

thenthen F FTT = m = m22 g – m g – m22aa

Substitute Substitute mm22 g – m g – m22a a in for in for FFTT in in Eqn 2Eqn 2..

And get And get mm11a = -ma = -m11 g + g + mm22 g – m g – m22a a

Simplify Simplify

mm11a = -ma = -m11 g + g + mm22 g – m g – m22aa

+ m+ m22a a + m + m22aa

mm11a a + m+ m22a a = -m= -m11 g + g + mm22 g g Factor out the a’s and g’sFactor out the a’s and g’s a(a(mm11 + m+ m22) = g() = g(-m-m11 + + mm22))

Solve for the accelerationSolve for the acceleration

a (a (mm11 + m+ m22) = g() = g(-m-m11 + + mm22))

((mm11 + m+ m22) () (mm11 + m+ m22) )

Solve for the accelerationSolve for the acceleration

a = g (a = g (-m-m11 + + mm22)) ((mm11 + m+ m22))

OROR

a = g (a = g (mm22 - m- m11)) ((mm11 + m+ m22))

ThereforeTherefore

a = g (a = g (mm22 - m- m11)) ((mm11 + m+ m22))

Therefore we can compare our Therefore we can compare our calculation from the above equation calculation from the above equation to the actual acceleration from our to the actual acceleration from our graph to determine the validity of graph to determine the validity of Newton’s 2Newton’s 2ndnd Law! Law!