Verification of Newton’s Second Law of Motion. presented by redwanul.pptx
Verification of Newton’s Second Law of
Motion by Atwood Machine
SL Name ID
1 FARIA, MAISHA 22-48478-3
2 SULTANA, SHARMIN 22-48479-3
3 MAZUMDER, REDWANUL HAQUE 22-48481-3
4 RAHMAN, FAHMIDA 22-48483-3
The net force(𝐹𝑛𝑒𝑡) on a body is equal to the product of the body’s mass (m) and its
acceleration (𝑎). In vector equation form, 𝐹𝑛𝑒𝑡 = 𝑚𝑎
Newton's second law can be formally stated as, the acceleration of an object as produced
by a net force is directly proportional to the magnitude of the net force, in the same
direction as the net force, and inversely proportional to the mass of the object. This
statement is expressed in equation form as, a = 𝐹𝑛𝑒𝑡 m .
Newton’s second law
Considering the upward direction as positive, neglecting frication and mass of
the pulley, and applying Newton’s law of motion we get
for M: 𝐹𝑛𝑒𝑡 = 𝑀𝑔 − 𝑇 = 𝑀𝑎,
for m: 𝐹𝑛𝑒𝑡 = T − mg = ma
Solving these two equations, we get the theoretical acceleration as
(𝑀 − 𝑚)
As acceleration due to gravity g is constant in a particular place and taking
total mass (M + m) constant for the Atwood machine, according to Newton’s
second law we get
𝑎𝑡ℎ ∝ (M – m)
According to fig. applying the knowledge of equations of motion
(D=ut +1/2 a𝑡2
),we can calculate the experimental acceleration by
The lighter mass (m) was held on the floor while attached to an end of the string. At
height (D) from the floor, the heavier mass (M) was attached to the other end of the
string. The height (D) was measured with a measuring scale.
The string was ran over the pulley in the vertical plain. The whole system was then
released. The time for the heavier mass to fall onto the floor was measured.
The experiment was ran for 7 different mass differences (M-m). It was made sure
that (𝑀 + 𝑚 ≈ 700𝑔) was constant.
Excel was used to plot graph of experimental data (a exp VS M-m) and the slope was
found from the best fit line. A trend line was added to the graph. Slope was set
(g/M-m) and was solved for (M + m). The percentage difference in (M + m) was
Graph of 𝒂𝒆𝒙 vs (M-m)
y = 1.3499x - 39.439
0 50 100 150 200 250 300 350
difference of mass,(M-m)(gm)
aex VS (M-m)graph
Analysis and Calculation
The slop of the straight line:
From the graph-
Or, M+m =
Error of :
× 100 = 3.712%
From the ‘acceleration vs mass difference’ graph, the relationship between
experimental acceleration and mass difference is proportional linear for the
Atwood machine same as the theory says. Thus we can say that newton’s
second law is 725.98gm.
Readings might not have been taken to eye level here measuring height.
Reaction time while taking reading from she-watch might have affected our result.
Take minimum 3 reading of time from a stop-watch and then calculate mean value
to minimize errors.
As there are lots of variables, calculation should be done carefully.
The string wasn't moving freely because of some frictional Problem.
Where, aex = experimental acceleration, D is the distance traveled by the object during time t.
If 𝑀 > 𝑚, the acceleration, a, with which the whole system moves is given by,
where, ath = theoretical acceleration and (g) is the acceleration due to gravity (9.80 m/s2 ).