Tram Huynh 210092 | 12K

ATWOOD’S MACHINE Aim: To determine the relationships between the masses on an Atwood’s machine and the acceleration.

Apparatus: As on prac sheet

Methods: As on prac sheet

Results & Calculations:

Data table Part I: Total Mass Constant

Trial m1 m2 Acceleration m mT(g) (g) (m/s2) (g) (g)

1 200 200 0 0 400

2 195 205 0.197 10 400

3 190 210 0.428 20 400

4 185 215 0.678 30 400

5 180 220 0.901 40 400

0 5 10 15 20 25 30 35 40 450

0.10.20.30.40.50.60.70.80.9

1

f(x) = 0.02283 x − 0.0158R² = 0.998570949055359

Part I: Acceleration vs. Mass Difference

m (g)

Acce

lera

tion

(ms-

2)

200 220 240 260 280 300 320 340 360 380 4000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

f(x) = − 0.002105 x + 1.2333R² = 0.962366700692831

Part II: Acceleration vs. Mass Total

Mass Total (g)

Acce

lera

tion

(ms-

2)

Data table Part II: The Mass Difference Constant

Trial m1 m2 Acceleration m mT(g) (g) (m/s2) (g) (g)

1 120 100 0.801 20 220

2 140 120 0.665 20 260

3 160 140 0.577 20 300

4 180 160 0.507 20 340

5 200 180 0.459 20 280

Tram Huynh 210092 | 12KThe mass difference was calculated by subtracting Mass 2 from Mass 1. For example, the mass difference of trial 1 of

Part 1 is found by Mass 1 – Mass 2 = 200g – 200g = 0 grams. The total mass was found by adding the two masses. For

example, for trial 1 of Part 2, the total mass is equal to 120g + 100g = 220 grams. All the calculations for mass

difference and total mass were calculated in the identical way.

Analysis:

Based on the Acceleration vs. Mass Difference graph generated from data table Part 1, it was found that the

relationship between acceleration and mass different was positively linear and a had a strong correlation value of

0.9986. Likewise, the Acceleration vs. Total Mass graph generated from data table Part 2 shows the relationship

between acceleration and total mass to be negatively linear with a strong correlation value of 0.9624.

Figure 1: Free body diagrams of mass 1 and mass 2 in an Atwood’s machine

Looking at Figure 1, mass 1 (m) shows that weight (mg) acts down and the tension, T, in the cord acts up. Since the

tension force is greater than the weight force, the net force acting on Mass 1 is equal to ∑ F=T−mg acting

upwards in the direction of its acceleration. Using Newton’s second law, ∑ F=ma, the equation ma=T−mg can

be concluded. Similarly, the net force acting Mass 2, M, is equal to ∑ F=Mg– T since the weight force is greater

than the tension in the rope, causing the mass to accelerate downwards. Again, using Newton’s second law,

Ma=Mg−T can be concluded.

Combining these two equations, the tension value will cancel out, resulting in ma+mg=Mg−Ma. Rearranging the

equation so that acceleration is on one side gives ma+Ma=Mg−mg. When acceleration is made the subject, then

the acceleration of an Atwood’s machine can be expressed aAcceleration=g∗(M−m)TotalMass

s.

Conclusion:

Tram Huynh 210092 | 12KBy measuring the acceleration of an Atwood’s machine, it was found that the total mass and the mass difference of

the two masses influence the acceleration. From the free body diagram, the expression for acceleration of the

system was found to be Acceleration=g∗(M−m)TotalMass

.