Exp 5: Dynamics
Objective: The objective of this experiment is to study Newton’s Second Law of
Motion utilizing the Atwood Machine and to show that the acceleration is proportional to
the force causing the motion.
Equipment:
• Personal computer
• Smart pulley
• Set of masses
• String
• Science workshop
• Universal clamp
• Digital balance
• 2 mass hangers
Theory: Newton’s Second Law of motion states that net force on the body is equal to the
product of the body’s mass and its acceleration:
r
r
F
∑ = m⋅a
The acceleration of the object depends on the net applied force and the mass.
Second Newton’s Law of motion can be proved by using Atwood’s machine. It consists
of two unequal masses connected by a string looped over a pulley.
The difference in weight between the hanging masses determines the net force acting on
the system. This net force accelerates both masses; the heavier mass is accelerated
downward and lighter- upward, assuming that pulley and string have no mass, that string
does not stretch and there is no friction between the pulley and the string.
Equation of motion for m1: m1 a = T − m1 g
Equation of motion for m2: m2 a = m2 g − T
where T-tension on the string
m1-lighter mass
m2-heavier mass
g-acceleration due to gravity
Those two equations of motion can be solved for a and T to get
m − m1
a= 2
g and
m1 + m2
2m1 m 2
T=
g
m1 + m 2
The net force of the system is
Fnet = (m1 + m2 )a = (m2 − m1 ) g
Data: Data was collected by dropping two connected, unequal masses m1 and m2 over a
pulley. We released the lighter mass m1 which was initially on the floor. As a result both
masses start to accelerate; m1 up and m2 down. The computer plotted V vs. t as m2 was
falling. The slope of this line was measured acceleration. We repeated procedure 3 more
time each time transferring 10g from m2 to m1 and keeping total mass of the system
constant.
Data table
m1±0.1 (g)
250.0
261.2
269.8
279.8
m2±0.1 (g)
319.7
309.2
299.3
289.3
ameas(m/s2)
1.14± 0.0016
0.780±0.0043
0.444±0.0026
0.120±.0.0039
\\
Calculations:
First I will calculate the net force using equation Fnet=(m2-m1)g for each run
Run 1: Fnet=(0.3197-0.2500)*9.79=0.682N
Run 2: Fnet=(0.3092-0.2612)*9.79=0.470N
Run 3: Fnet=(0.2993-0.2698)*9.79=0.289N
Run 4: Fnet=(0.2893-0.2798)*9.79=0.093N
Then I will calculate the total mass of the system from run 1:
mtot=m1+m2=0.250+0.3197=0.5697kg
The uncertainty is
δmtot=δm1+δm2=0.1+0.1=0.2g=0.0002kg
So, total mass of the system is mtot=0.5697±0.0002kg
Using JagFit, I will plot Fnet vs ameas and find the slope b
b=0.5738+/-0.008431kg
If the Newton’s second law is valid then b should be equal to mtot. To prove it, I will
compare b with mtot
mtot − b
δmtot + δb
=
0.5697 − 0.5738
= 0. 5
0.0002 + 0.008431
Conclusion: For this experiment I was testing Newton’s Second Law of motion by
calculating the net force of the system of 2 unequal masses and plot is as a function of
acceleration measured by Data Studio. The slope returned from this graph was
b=0.574+/-0.00843kg and the actual total mass of the system was
mtot=0.5697±0.0002kg.When compared those two were within 3 standard devotions (0.5)
from each other so we were able to prove Newton’s Second Law of motion. Possible
sources of error were the fact that we neglected masses of the pulley and the string,
friction between them and we did not take into account moment of inertia of the pulley
and stretching of the string.