LAB RECORD FOR THE Verification of newton's law

1. CENTURION UNIVERSITY OF TECHNOLOGY AND MANAGEMNT
MECHANICS FOR ENGINEER
TOPIC: VERIFICATION OF NEWTON’S LAW OF MOTION
NAME: - KALINGO AUROBINDO NAYAK
SEC- D
REG NO-200101120022
BRANCH: -CSE
CAMPUS: -PARALAKHEMUNDI

2. AIM :
• To find the acceleration of the cart in the simulator.
• To find the distance covered by the cart in the simulator in the given time
interval.
Theory
Consider a cart on a low-friction track as shown in Fig. 1. A light string is attached
to the cart and passes over a pulley at the end of the track and a second mass is
attached to the end of this string. The weight of the hanging mass provides
tension in the string, which helps to accelerate the cart along the track. A small
frictional force will resist this motion. We assume that the string is massless (or of
negligible mass) and there is no friction between the string and the pulley.
Therefore, the tension in the string will be the same at all points along the string.
This results in both masses having the same magnitude of acceleration but the
direction of the acceleration will be different. The cart will accelerate to the right
while hanging mass will accelerate in the downward direction as shown in Fig. 1.
We will take the positive direction to be in the direction of the acceleration of the
two masses as indicated by the coordinate axes system in Fig 1.
The free-body diagrams for the two masses are shown in Fig. 2. Let's look at the
forces acting on each mass.

3. For the falling mass m, there are no forces acting in the horizontal direction. In the vertical direction it is
pulled downward by gravity giving the object weight, W = mg and upward by the tension T in the string.
See Fig. 2b. Thus, Newton's second law applied to the falling mass in the y direction will be
mg-T=ma ….[1]
Figure 2a shows the forces acting on M. There is no motion of the cart in the vertical direction.
Therefore, the net force in the vertical direction will be zero, as will the acceleration. In the horizontal
direction, the tension in the string acts in the +x direction on the cart while the friction force between
the cart's tires and the surface of the track acts in the –x direction. Newton's second law, in the x and y
directions, respectively, are
T-f=Ma …. [2]
R-Mg=0 ….[3]
From equation [3], we have R=Mg
And f=force of friction=μR=μMg
Hence equation [2] gives T- μMg =Ma …..[4]
Adding eq [1] and eq [4],
we get mg- μMg=(M+m)a
or, a= (mg- μMg)/(M+m) …… [5]
.

4. Observations
Sl.no m in g M in g μ a=(mg-
μMg)/(
M+m )
=x
S in m t in s a=2S/t2
=y
x-y
1 1 10 0.002 0.873 1 1.514 0.873 ---
2 2 20 0.002 0.873 1 1.514 0.873 ---
3 16 30 0.002 3.395 1 0.767 3.396 0.001
4 25 40 0.002 3.757 1 0.730 3.753 0.004
5 33 50 0.003 3.878 1 0.718 3.883 0.005
6 37 50 0.003 4.151 1 0.694 4.158 0.007
7 40 60 0.004 3.896 0.5 0.512 3.893 0.003
8 43 70 0.005 3.699 0.5 0.525 3.710 0.011
9 46 90 0.005 3.282 0.5 0.557 3.290 0.008
10 50 100 0.005 3.254 0.5 0.560 3.256 0.002